Compare nested objects in JavaScript and return keys equality





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6















I have two nested objects obj1 and obj2 and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag



So for a given obj1 like



obj1 = {

  prop1: 1,

  prop2: "foo",

  prop3: {

    prop4: 2,

    prop5: "bar" 

  }

}


and the obj2 like



obj2 = {

      prop1: 3,

      prop2: "foo",

      prop3: {

        prop4: 2,

        prop5: "foobar" 

      },

      prop6: "new"

    }


it should return



equality = {

     prop1: false,

     prop2: true,

     prop3 : {

       prop4: true,

       prop5: false

     },

     prop6: false

   }


If an object has a new property, like obj2.prop6, then the equality will be equality.prop6 = false.



For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties






javascript







share|improve this question




















  • 1





    Will both objects always have exact match properties?

    – holydragon
    3 hours ago











  • good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

    – loretoparisi
    3 hours ago




















6















I have two nested objects obj1 and obj2 and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag



So for a given obj1 like



obj1 = {

  prop1: 1,

  prop2: "foo",

  prop3: {

    prop4: 2,

    prop5: "bar" 

  }

}


and the obj2 like



obj2 = {

      prop1: 3,

      prop2: "foo",

      prop3: {

        prop4: 2,

        prop5: "foobar" 

      },

      prop6: "new"

    }


it should return



equality = {

     prop1: false,

     prop2: true,

     prop3 : {

       prop4: true,

       prop5: false

     },

     prop6: false

   }


If an object has a new property, like obj2.prop6, then the equality will be equality.prop6 = false.



For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties






javascript







share|improve this question




















  • 1





    Will both objects always have exact match properties?

    – holydragon
    3 hours ago











  • good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

    – loretoparisi
    3 hours ago
















6












6








6


1






I have two nested objects obj1 and obj2 and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag



So for a given obj1 like



obj1 = {

  prop1: 1,

  prop2: "foo",

  prop3: {

    prop4: 2,

    prop5: "bar" 

  }

}


and the obj2 like



obj2 = {

      prop1: 3,

      prop2: "foo",

      prop3: {

        prop4: 2,

        prop5: "foobar" 

      },

      prop6: "new"

    }


it should return



equality = {

     prop1: false,

     prop2: true,

     prop3 : {

       prop4: true,

       prop5: false

     },

     prop6: false

   }


If an object has a new property, like obj2.prop6, then the equality will be equality.prop6 = false.



For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties






javascript







share|improve this question
















I have two nested objects obj1 and obj2 and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag



So for a given obj1 like



obj1 = {

  prop1: 1,

  prop2: "foo",

  prop3: {

    prop4: 2,

    prop5: "bar" 

  }

}


and the obj2 like



obj2 = {

      prop1: 3,

      prop2: "foo",

      prop3: {

        prop4: 2,

        prop5: "foobar" 

      },

      prop6: "new"

    }


it should return



equality = {

     prop1: false,

     prop2: true,

     prop3 : {

       prop4: true,

       prop5: false

     },

     prop6: false

   }


If an object has a new property, like obj2.prop6, then the equality will be equality.prop6 = false.



For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties






javascript




javascript






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago







loretoparisi

















asked 3 hours ago









loretoparisiloretoparisi

8,06654973




8,06654973








  • 1





    Will both objects always have exact match properties?

    – holydragon
    3 hours ago











  • good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

    – loretoparisi
    3 hours ago
















  • 1





    Will both objects always have exact match properties?

    – holydragon
    3 hours ago











  • good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

    – loretoparisi
    3 hours ago










1




1





Will both objects always have exact match properties?

– holydragon
3 hours ago





Will both objects always have exact match properties?

– holydragon
3 hours ago













good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

– loretoparisi
3 hours ago







good point. Nope, so the equality could have a new key set to false. Updating with this point. Thank you.

– loretoparisi
3 hours ago














4 Answers
4






active

oldest

votes


















1














You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } }

const obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" } }



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r

    else return r[e] || undefined

  }, obj)

}



function compare(a, b, prev = "") {

  return Object.keys(a).reduce((r, e) => {

    const path = prev + (prev ? '.' + e : e);

    const value = a[e] === get(b, path);

    r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value

    return r;

  }, {})

}



const result = compare(obj1, obj2);

console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = {"prop1":1,"prop2":"foo","prop3":{"prop4":2,"prop5":"bar"},"prop7":{"prop9":{"prop10":"foo"}}}

const obj2 = {"prop1":3,"prop2":"foo","prop3":{"prop4":2,"prop5":"foobar"},"prop6":"new","prop7":{"foo":"foo","bar":{"baz":"baz"}}}



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r;

    else return r[e] || undefined;

  }, obj);

}



function isEmpty(o) {

  if (typeof o !== 'object') return true;

  else return !Object.keys(o).length;

}



function build(a, b, o = null, prev = '') {

  return Object.keys(a).reduce(

    (r, e) => {

      const path = prev + (prev ? '.' + e : e);

      const bObj = get(b, path);

      const value = a[e] === bObj;



      if (typeof a[e] === 'object') {

        if (isEmpty(a[e]) && isEmpty(bObj)) {

          if (e in r) r[e] = r[e];

          else r[e] = true;

        } else if (!bObj && isEmpty(a[e])) {

          r[e] = value;

        } else {

          r[e] = build(a[e], b, r[e], path);

        }

      } else {

        r[e] = value;

      }

      return r;

    },

    o ? o : {}

  );

}



function compare(a, b) {

  const o = build(a, b);

  return build(b, a, o);

}



const result = compare(obj1, obj2);

console.log(result)








share|improve this answer





















  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 hours ago











  • Just a thing, supposed that the property is a void object like {}, it does not make the comparison.

    – loretoparisi
    2 hours ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    1 hour ago











  • Ok, now it handles a nested lonely prop11={} object, but if you have like the same prop12={} in both, you will get the {} as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    1 hour ago



















4














You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';



function getDifference(a, b) {

    return Object.assign(...Array.from(

        new Set([...Object.keys(a), ...Object.keys(b)]),

        k => ({ [k]: isObject(a[k]) && isObject(b[k])

            ? getDifference(a[k], b[k])

            : a[k] === b[k]

        })

    ));

}



var obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } },

    obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" }, prop6: "new" };



console.log(getDifference(obj1, obj2));








share|improve this answer


























  • Thanks! This solution works in every condition even if the property is a void object like prop7 = {}. Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 hours ago








  • 1





    without spread not Set.

    – Nina Scholz
    2 hours ago











  • maybe like Object.assign(Array.from(new Set( .concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    1 hour ago











  • w h a t e v e r ...

    – Nina Scholz
    1 hour ago



















2














Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || {} check is added.






const obj1={prop1:1,prop2:"foo",prop3:{prop4:2,prop5:"bar"},prop7:{pro8:"only in 1"}},

    obj2={prop1:3,prop2:"foo",prop3:{prop4:2,prop5:"foobar"}, prop6: "only in 2"};

    

const isObject = val => val && typeof val === 'object'; // required for "null" comparison



function compare(obj1, obj2) {

  let equality = {},

      merged = { ...obj1, ...obj2 }; // has properties of both



  for (let key in merged) {

    let value1 = obj1[key], value2 = obj2[key];



    if (isObject(value1) || isObject(value2)) {

      equality[key] = compare(value1 || {}, value2 || {});

    } else {

      equality[key] = value1 === value2

    }

  }



  return equality;

}



console.log(compare(obj1, obj2))








share|improve this answer


























  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    3 hours ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    3 hours ago











  • @loretoparisi updated

    – adiga
    3 hours ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    3 hours ago













  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    3 hours ago



















1














A recursive example,






var obj1 = {

        prop1: 1,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "bar"

        },

        prop7: {},

    }



    var obj2 = {

        prop1: 3,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "foobar"

        },

        prop6: "new",

        prop7: {},

        prop8: {},

    }



    var result = {};



    function compare(obj1, obj2, obj_) {

        for (let k in obj1) {

            var type = typeof obj1[k];

            if (type === 'object') {

                obj_[k] = {};

                if (!obj2[k]){

                    obj_[k] = false;

                }else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object)) {

                    obj_[k] = true;

                } else {

                    compare(obj1[k], obj2[k], obj_[k]);

                }

            } else {

                obj_[k] = (obj1[k] === obj2[k]);

            }



        }

    }



    if (Object.keys(obj1).length < Object.keys(obj2).length) { //check if both objects vary in length.

        var tmp = obj1;

        obj1 = obj2;

        obj2 = tmp;

    }



    compare(obj1, obj2, result);



    console.log(result);








share|improve this answer


























  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a {} in the results for properties of type object that belong to both objects and that are {}.

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    58 mins ago












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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } }

const obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" } }



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r

    else return r[e] || undefined

  }, obj)

}



function compare(a, b, prev = "") {

  return Object.keys(a).reduce((r, e) => {

    const path = prev + (prev ? '.' + e : e);

    const value = a[e] === get(b, path);

    r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value

    return r;

  }, {})

}



const result = compare(obj1, obj2);

console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = {"prop1":1,"prop2":"foo","prop3":{"prop4":2,"prop5":"bar"},"prop7":{"prop9":{"prop10":"foo"}}}

const obj2 = {"prop1":3,"prop2":"foo","prop3":{"prop4":2,"prop5":"foobar"},"prop6":"new","prop7":{"foo":"foo","bar":{"baz":"baz"}}}



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r;

    else return r[e] || undefined;

  }, obj);

}



function isEmpty(o) {

  if (typeof o !== 'object') return true;

  else return !Object.keys(o).length;

}



function build(a, b, o = null, prev = '') {

  return Object.keys(a).reduce(

    (r, e) => {

      const path = prev + (prev ? '.' + e : e);

      const bObj = get(b, path);

      const value = a[e] === bObj;



      if (typeof a[e] === 'object') {

        if (isEmpty(a[e]) && isEmpty(bObj)) {

          if (e in r) r[e] = r[e];

          else r[e] = true;

        } else if (!bObj && isEmpty(a[e])) {

          r[e] = value;

        } else {

          r[e] = build(a[e], b, r[e], path);

        }

      } else {

        r[e] = value;

      }

      return r;

    },

    o ? o : {}

  );

}



function compare(a, b) {

  const o = build(a, b);

  return build(b, a, o);

}



const result = compare(obj1, obj2);

console.log(result)








share|improve this answer





















  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 hours ago











  • Just a thing, supposed that the property is a void object like {}, it does not make the comparison.

    – loretoparisi
    2 hours ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    1 hour ago











  • Ok, now it handles a nested lonely prop11={} object, but if you have like the same prop12={} in both, you will get the {} as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    1 hour ago
















1














You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } }

const obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" } }



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r

    else return r[e] || undefined

  }, obj)

}



function compare(a, b, prev = "") {

  return Object.keys(a).reduce((r, e) => {

    const path = prev + (prev ? '.' + e : e);

    const value = a[e] === get(b, path);

    r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value

    return r;

  }, {})

}



const result = compare(obj1, obj2);

console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = {"prop1":1,"prop2":"foo","prop3":{"prop4":2,"prop5":"bar"},"prop7":{"prop9":{"prop10":"foo"}}}

const obj2 = {"prop1":3,"prop2":"foo","prop3":{"prop4":2,"prop5":"foobar"},"prop6":"new","prop7":{"foo":"foo","bar":{"baz":"baz"}}}



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r;

    else return r[e] || undefined;

  }, obj);

}



function isEmpty(o) {

  if (typeof o !== 'object') return true;

  else return !Object.keys(o).length;

}



function build(a, b, o = null, prev = '') {

  return Object.keys(a).reduce(

    (r, e) => {

      const path = prev + (prev ? '.' + e : e);

      const bObj = get(b, path);

      const value = a[e] === bObj;



      if (typeof a[e] === 'object') {

        if (isEmpty(a[e]) && isEmpty(bObj)) {

          if (e in r) r[e] = r[e];

          else r[e] = true;

        } else if (!bObj && isEmpty(a[e])) {

          r[e] = value;

        } else {

          r[e] = build(a[e], b, r[e], path);

        }

      } else {

        r[e] = value;

      }

      return r;

    },

    o ? o : {}

  );

}



function compare(a, b) {

  const o = build(a, b);

  return build(b, a, o);

}



const result = compare(obj1, obj2);

console.log(result)








share|improve this answer





















  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 hours ago











  • Just a thing, supposed that the property is a void object like {}, it does not make the comparison.

    – loretoparisi
    2 hours ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    1 hour ago











  • Ok, now it handles a nested lonely prop11={} object, but if you have like the same prop12={} in both, you will get the {} as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    1 hour ago














1












1








1







You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } }

const obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" } }



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r

    else return r[e] || undefined

  }, obj)

}



function compare(a, b, prev = "") {

  return Object.keys(a).reduce((r, e) => {

    const path = prev + (prev ? '.' + e : e);

    const value = a[e] === get(b, path);

    r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value

    return r;

  }, {})

}



const result = compare(obj1, obj2);

console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = {"prop1":1,"prop2":"foo","prop3":{"prop4":2,"prop5":"bar"},"prop7":{"prop9":{"prop10":"foo"}}}

const obj2 = {"prop1":3,"prop2":"foo","prop3":{"prop4":2,"prop5":"foobar"},"prop6":"new","prop7":{"foo":"foo","bar":{"baz":"baz"}}}



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r;

    else return r[e] || undefined;

  }, obj);

}



function isEmpty(o) {

  if (typeof o !== 'object') return true;

  else return !Object.keys(o).length;

}



function build(a, b, o = null, prev = '') {

  return Object.keys(a).reduce(

    (r, e) => {

      const path = prev + (prev ? '.' + e : e);

      const bObj = get(b, path);

      const value = a[e] === bObj;



      if (typeof a[e] === 'object') {

        if (isEmpty(a[e]) && isEmpty(bObj)) {

          if (e in r) r[e] = r[e];

          else r[e] = true;

        } else if (!bObj && isEmpty(a[e])) {

          r[e] = value;

        } else {

          r[e] = build(a[e], b, r[e], path);

        }

      } else {

        r[e] = value;

      }

      return r;

    },

    o ? o : {}

  );

}



function compare(a, b) {

  const o = build(a, b);

  return build(b, a, o);

}



const result = compare(obj1, obj2);

console.log(result)








share|improve this answer















You could use reduce to build new object and another get method to get nested props from other object by string and compare it to current prop value in first object.






const obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } }

const obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" } }



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r

    else return r[e] || undefined

  }, obj)

}



function compare(a, b, prev = "") {

  return Object.keys(a).reduce((r, e) => {

    const path = prev + (prev ? '.' + e : e);

    const value = a[e] === get(b, path);

    r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value

    return r;

  }, {})

}



const result = compare(obj1, obj2);

console.log(result)





To compare all properties of both objects you could create extra function that will perform loop by both objects.






const obj1 = {"prop1":1,"prop2":"foo","prop3":{"prop4":2,"prop5":"bar"},"prop7":{"prop9":{"prop10":"foo"}}}

const obj2 = {"prop1":3,"prop2":"foo","prop3":{"prop4":2,"prop5":"foobar"},"prop6":"new","prop7":{"foo":"foo","bar":{"baz":"baz"}}}



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r;

    else return r[e] || undefined;

  }, obj);

}



function isEmpty(o) {

  if (typeof o !== 'object') return true;

  else return !Object.keys(o).length;

}



function build(a, b, o = null, prev = '') {

  return Object.keys(a).reduce(

    (r, e) => {

      const path = prev + (prev ? '.' + e : e);

      const bObj = get(b, path);

      const value = a[e] === bObj;



      if (typeof a[e] === 'object') {

        if (isEmpty(a[e]) && isEmpty(bObj)) {

          if (e in r) r[e] = r[e];

          else r[e] = true;

        } else if (!bObj && isEmpty(a[e])) {

          r[e] = value;

        } else {

          r[e] = build(a[e], b, r[e], path);

        }

      } else {

        r[e] = value;

      }

      return r;

    },

    o ? o : {}

  );

}



function compare(a, b) {

  const o = build(a, b);

  return build(b, a, o);

}



const result = compare(obj1, obj2);

console.log(result)








const obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } }

const obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" } }



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r

    else return r[e] || undefined

  }, obj)

}



function compare(a, b, prev = "") {

  return Object.keys(a).reduce((r, e) => {

    const path = prev + (prev ? '.' + e : e);

    const value = a[e] === get(b, path);

    r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value

    return r;

  }, {})

}



const result = compare(obj1, obj2);

console.log(result)





const obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } }

const obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" } }



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r

    else return r[e] || undefined

  }, obj)

}



function compare(a, b, prev = "") {

  return Object.keys(a).reduce((r, e) => {

    const path = prev + (prev ? '.' + e : e);

    const value = a[e] === get(b, path);

    r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value

    return r;

  }, {})

}



const result = compare(obj1, obj2);

console.log(result)





const obj1 = {"prop1":1,"prop2":"foo","prop3":{"prop4":2,"prop5":"bar"},"prop7":{"prop9":{"prop10":"foo"}}}

const obj2 = {"prop1":3,"prop2":"foo","prop3":{"prop4":2,"prop5":"foobar"},"prop6":"new","prop7":{"foo":"foo","bar":{"baz":"baz"}}}



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r;

    else return r[e] || undefined;

  }, obj);

}



function isEmpty(o) {

  if (typeof o !== 'object') return true;

  else return !Object.keys(o).length;

}



function build(a, b, o = null, prev = '') {

  return Object.keys(a).reduce(

    (r, e) => {

      const path = prev + (prev ? '.' + e : e);

      const bObj = get(b, path);

      const value = a[e] === bObj;



      if (typeof a[e] === 'object') {

        if (isEmpty(a[e]) && isEmpty(bObj)) {

          if (e in r) r[e] = r[e];

          else r[e] = true;

        } else if (!bObj && isEmpty(a[e])) {

          r[e] = value;

        } else {

          r[e] = build(a[e], b, r[e], path);

        }

      } else {

        r[e] = value;

      }

      return r;

    },

    o ? o : {}

  );

}



function compare(a, b) {

  const o = build(a, b);

  return build(b, a, o);

}



const result = compare(obj1, obj2);

console.log(result)





const obj1 = {"prop1":1,"prop2":"foo","prop3":{"prop4":2,"prop5":"bar"},"prop7":{"prop9":{"prop10":"foo"}}}

const obj2 = {"prop1":3,"prop2":"foo","prop3":{"prop4":2,"prop5":"foobar"},"prop6":"new","prop7":{"foo":"foo","bar":{"baz":"baz"}}}



function get(obj, path) {

  return path.split('.').reduce((r, e) => {

    if (!r) return r;

    else return r[e] || undefined;

  }, obj);

}



function isEmpty(o) {

  if (typeof o !== 'object') return true;

  else return !Object.keys(o).length;

}



function build(a, b, o = null, prev = '') {

  return Object.keys(a).reduce(

    (r, e) => {

      const path = prev + (prev ? '.' + e : e);

      const bObj = get(b, path);

      const value = a[e] === bObj;



      if (typeof a[e] === 'object') {

        if (isEmpty(a[e]) && isEmpty(bObj)) {

          if (e in r) r[e] = r[e];

          else r[e] = true;

        } else if (!bObj && isEmpty(a[e])) {

          r[e] = value;

        } else {

          r[e] = build(a[e], b, r[e], path);

        }

      } else {

        r[e] = value;

      }

      return r;

    },

    o ? o : {}

  );

}



function compare(a, b) {

  const o = build(a, b);

  return build(b, a, o);

}



const result = compare(obj1, obj2);

console.log(result)






share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 3 hours ago









Nenad VracarNenad Vracar

73.3k126085




73.3k126085








  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 hours ago











  • Just a thing, supposed that the property is a void object like {}, it does not make the comparison.

    – loretoparisi
    2 hours ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    1 hour ago











  • Ok, now it handles a nested lonely prop11={} object, but if you have like the same prop12={} in both, you will get the {} as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    1 hour ago














  • 1





    I thinks this is the best solution since it supports all recent version of ECMAScript.

    – loretoparisi
    2 hours ago











  • Just a thing, supposed that the property is a void object like {}, it does not make the comparison.

    – loretoparisi
    2 hours ago






  • 1





    In that case you could use this jsfiddle.net/vqwn3zLf

    – Nenad Vracar
    1 hour ago











  • Ok, now it handles a nested lonely prop11={} object, but if you have like the same prop12={} in both, you will get the {} as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi Try this jsfiddle.net/r0y8nd3q/1

    – Nenad Vracar
    1 hour ago








1




1





I thinks this is the best solution since it supports all recent version of ECMAScript.

– loretoparisi
2 hours ago





I thinks this is the best solution since it supports all recent version of ECMAScript.

– loretoparisi
2 hours ago













Just a thing, supposed that the property is a void object like {}, it does not make the comparison.

– loretoparisi
2 hours ago





Just a thing, supposed that the property is a void object like {}, it does not make the comparison.

– loretoparisi
2 hours ago




1




1





In that case you could use this jsfiddle.net/vqwn3zLf

– Nenad Vracar
1 hour ago





In that case you could use this jsfiddle.net/vqwn3zLf

– Nenad Vracar
1 hour ago













Ok, now it handles a nested lonely prop11={} object, but if you have like the same prop12={} in both, you will get the {} as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

– loretoparisi
1 hour ago







Ok, now it handles a nested lonely prop11={} object, but if you have like the same prop12={} in both, you will get the {} as result for the prop12 keyword, instead of bool. See here jsfiddle.net/gpu20nwy

– loretoparisi
1 hour ago






1




1





@loretoparisi Try this jsfiddle.net/r0y8nd3q/1

– Nenad Vracar
1 hour ago





@loretoparisi Try this jsfiddle.net/r0y8nd3q/1

– Nenad Vracar
1 hour ago













4














You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';



function getDifference(a, b) {

    return Object.assign(...Array.from(

        new Set([...Object.keys(a), ...Object.keys(b)]),

        k => ({ [k]: isObject(a[k]) && isObject(b[k])

            ? getDifference(a[k], b[k])

            : a[k] === b[k]

        })

    ));

}



var obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } },

    obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" }, prop6: "new" };



console.log(getDifference(obj1, obj2));








share|improve this answer


























  • Thanks! This solution works in every condition even if the property is a void object like prop7 = {}. Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 hours ago








  • 1





    without spread not Set.

    – Nina Scholz
    2 hours ago











  • maybe like Object.assign(Array.from(new Set( .concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    1 hour ago











  • w h a t e v e r ...

    – Nina Scholz
    1 hour ago
















4














You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';



function getDifference(a, b) {

    return Object.assign(...Array.from(

        new Set([...Object.keys(a), ...Object.keys(b)]),

        k => ({ [k]: isObject(a[k]) && isObject(b[k])

            ? getDifference(a[k], b[k])

            : a[k] === b[k]

        })

    ));

}



var obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } },

    obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" }, prop6: "new" };



console.log(getDifference(obj1, obj2));








share|improve this answer


























  • Thanks! This solution works in every condition even if the property is a void object like prop7 = {}. Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 hours ago








  • 1





    without spread not Set.

    – Nina Scholz
    2 hours ago











  • maybe like Object.assign(Array.from(new Set( .concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    1 hour ago











  • w h a t e v e r ...

    – Nina Scholz
    1 hour ago














4












4








4







You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';



function getDifference(a, b) {

    return Object.assign(...Array.from(

        new Set([...Object.keys(a), ...Object.keys(b)]),

        k => ({ [k]: isObject(a[k]) && isObject(b[k])

            ? getDifference(a[k], b[k])

            : a[k] === b[k]

        })

    ));

}



var obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } },

    obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" }, prop6: "new" };



console.log(getDifference(obj1, obj2));








share|improve this answer















You could iterate all keys and check the nested objects if both values are objects.






const isObject = v => v && typeof v === 'object';



function getDifference(a, b) {

    return Object.assign(...Array.from(

        new Set([...Object.keys(a), ...Object.keys(b)]),

        k => ({ [k]: isObject(a[k]) && isObject(b[k])

            ? getDifference(a[k], b[k])

            : a[k] === b[k]

        })

    ));

}



var obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } },

    obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" }, prop6: "new" };



console.log(getDifference(obj1, obj2));








const isObject = v => v && typeof v === 'object';



function getDifference(a, b) {

    return Object.assign(...Array.from(

        new Set([...Object.keys(a), ...Object.keys(b)]),

        k => ({ [k]: isObject(a[k]) && isObject(b[k])

            ? getDifference(a[k], b[k])

            : a[k] === b[k]

        })

    ));

}



var obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } },

    obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" }, prop6: "new" };



console.log(getDifference(obj1, obj2));





const isObject = v => v && typeof v === 'object';



function getDifference(a, b) {

    return Object.assign(...Array.from(

        new Set([...Object.keys(a), ...Object.keys(b)]),

        k => ({ [k]: isObject(a[k]) && isObject(b[k])

            ? getDifference(a[k], b[k])

            : a[k] === b[k]

        })

    ));

}



var obj1 = { prop1: 1, prop2: "foo", prop3: { prop4: 2, prop5: "bar" } },

    obj2 = { prop1: 3, prop2: "foo", prop3: { prop4: 2, prop5: "foobar" }, prop6: "new" };



console.log(getDifference(obj1, obj2));






share|improve this answer














share|improve this answer



share|improve this answer








edited 2 hours ago

























answered 3 hours ago









Nina ScholzNina Scholz

197k15109179




197k15109179













  • Thanks! This solution works in every condition even if the property is a void object like prop7 = {}. Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 hours ago








  • 1





    without spread not Set.

    – Nina Scholz
    2 hours ago











  • maybe like Object.assign(Array.from(new Set( .concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    1 hour ago











  • w h a t e v e r ...

    – Nina Scholz
    1 hour ago



















  • Thanks! This solution works in every condition even if the property is a void object like prop7 = {}. Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

    – loretoparisi
    2 hours ago








  • 1





    without spread not Set.

    – Nina Scholz
    2 hours ago











  • maybe like Object.assign(Array.from(new Set( .concat(Object.keys(a)).concat(Object.keys(b)))

    – loretoparisi
    1 hour ago











  • w h a t e v e r ...

    – Nina Scholz
    1 hour ago

















Thanks! This solution works in every condition even if the property is a void object like prop7 = {}. Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

– loretoparisi
2 hours ago







Thanks! This solution works in every condition even if the property is a void object like prop7 = {}. Maybe avoiding the ECMA6 Spread notation for more compatibility could help.

– loretoparisi
2 hours ago






1




1





without spread not Set.

– Nina Scholz
2 hours ago





without spread not Set.

– Nina Scholz
2 hours ago













maybe like Object.assign(Array.from(new Set( .concat(Object.keys(a)).concat(Object.keys(b)))

– loretoparisi
1 hour ago





maybe like Object.assign(Array.from(new Set( .concat(Object.keys(a)).concat(Object.keys(b)))

– loretoparisi
1 hour ago













w h a t e v e r ...

– Nina Scholz
1 hour ago





w h a t e v e r ...

– Nina Scholz
1 hour ago











2














Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || {} check is added.






const obj1={prop1:1,prop2:"foo",prop3:{prop4:2,prop5:"bar"},prop7:{pro8:"only in 1"}},

    obj2={prop1:3,prop2:"foo",prop3:{prop4:2,prop5:"foobar"}, prop6: "only in 2"};

    

const isObject = val => val && typeof val === 'object'; // required for "null" comparison



function compare(obj1, obj2) {

  let equality = {},

      merged = { ...obj1, ...obj2 }; // has properties of both



  for (let key in merged) {

    let value1 = obj1[key], value2 = obj2[key];



    if (isObject(value1) || isObject(value2)) {

      equality[key] = compare(value1 || {}, value2 || {});

    } else {

      equality[key] = value1 === value2

    }

  }



  return equality;

}



console.log(compare(obj1, obj2))








share|improve this answer


























  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    3 hours ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    3 hours ago











  • @loretoparisi updated

    – adiga
    3 hours ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    3 hours ago













  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    3 hours ago
















2














Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || {} check is added.






const obj1={prop1:1,prop2:"foo",prop3:{prop4:2,prop5:"bar"},prop7:{pro8:"only in 1"}},

    obj2={prop1:3,prop2:"foo",prop3:{prop4:2,prop5:"foobar"}, prop6: "only in 2"};

    

const isObject = val => val && typeof val === 'object'; // required for "null" comparison



function compare(obj1, obj2) {

  let equality = {},

      merged = { ...obj1, ...obj2 }; // has properties of both



  for (let key in merged) {

    let value1 = obj1[key], value2 = obj2[key];



    if (isObject(value1) || isObject(value2)) {

      equality[key] = compare(value1 || {}, value2 || {});

    } else {

      equality[key] = value1 === value2

    }

  }



  return equality;

}



console.log(compare(obj1, obj2))








share|improve this answer


























  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    3 hours ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    3 hours ago











  • @loretoparisi updated

    – adiga
    3 hours ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    3 hours ago













  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    3 hours ago














2












2








2







Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || {} check is added.






const obj1={prop1:1,prop2:"foo",prop3:{prop4:2,prop5:"bar"},prop7:{pro8:"only in 1"}},

    obj2={prop1:3,prop2:"foo",prop3:{prop4:2,prop5:"foobar"}, prop6: "only in 2"};

    

const isObject = val => val && typeof val === 'object'; // required for "null" comparison



function compare(obj1, obj2) {

  let equality = {},

      merged = { ...obj1, ...obj2 }; // has properties of both



  for (let key in merged) {

    let value1 = obj1[key], value2 = obj2[key];



    if (isObject(value1) || isObject(value2)) {

      equality[key] = compare(value1 || {}, value2 || {});

    } else {

      equality[key] = value1 === value2

    }

  }



  return equality;

}



console.log(compare(obj1, obj2))








share|improve this answer















Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value || {} check is added.






const obj1={prop1:1,prop2:"foo",prop3:{prop4:2,prop5:"bar"},prop7:{pro8:"only in 1"}},

    obj2={prop1:3,prop2:"foo",prop3:{prop4:2,prop5:"foobar"}, prop6: "only in 2"};

    

const isObject = val => val && typeof val === 'object'; // required for "null" comparison



function compare(obj1, obj2) {

  let equality = {},

      merged = { ...obj1, ...obj2 }; // has properties of both



  for (let key in merged) {

    let value1 = obj1[key], value2 = obj2[key];



    if (isObject(value1) || isObject(value2)) {

      equality[key] = compare(value1 || {}, value2 || {});

    } else {

      equality[key] = value1 === value2

    }

  }



  return equality;

}



console.log(compare(obj1, obj2))








const obj1={prop1:1,prop2:"foo",prop3:{prop4:2,prop5:"bar"},prop7:{pro8:"only in 1"}},

    obj2={prop1:3,prop2:"foo",prop3:{prop4:2,prop5:"foobar"}, prop6: "only in 2"};

    

const isObject = val => val && typeof val === 'object'; // required for "null" comparison



function compare(obj1, obj2) {

  let equality = {},

      merged = { ...obj1, ...obj2 }; // has properties of both



  for (let key in merged) {

    let value1 = obj1[key], value2 = obj2[key];



    if (isObject(value1) || isObject(value2)) {

      equality[key] = compare(value1 || {}, value2 || {});

    } else {

      equality[key] = value1 === value2

    }

  }



  return equality;

}



console.log(compare(obj1, obj2))





const obj1={prop1:1,prop2:"foo",prop3:{prop4:2,prop5:"bar"},prop7:{pro8:"only in 1"}},

    obj2={prop1:3,prop2:"foo",prop3:{prop4:2,prop5:"foobar"}, prop6: "only in 2"};

    

const isObject = val => val && typeof val === 'object'; // required for "null" comparison



function compare(obj1, obj2) {

  let equality = {},

      merged = { ...obj1, ...obj2 }; // has properties of both



  for (let key in merged) {

    let value1 = obj1[key], value2 = obj2[key];



    if (isObject(value1) || isObject(value2)) {

      equality[key] = compare(value1 || {}, value2 || {});

    } else {

      equality[key] = value1 === value2

    }

  }



  return equality;

}



console.log(compare(obj1, obj2))






share|improve this answer














share|improve this answer



share|improve this answer








edited 2 hours ago

























answered 3 hours ago









adigaadiga

12.3k62645




12.3k62645













  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    3 hours ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    3 hours ago











  • @loretoparisi updated

    – adiga
    3 hours ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    3 hours ago













  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    3 hours ago



















  • Thanks, just updated the code, assuming you can have new properties as well.

    – loretoparisi
    3 hours ago











  • what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

    – AZ_
    3 hours ago











  • @loretoparisi updated

    – adiga
    3 hours ago











  • @adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

    – AZ_
    3 hours ago













  • @AZ_ updated. Not sure if it will fail for any scenario

    – adiga
    3 hours ago

















Thanks, just updated the code, assuming you can have new properties as well.

– loretoparisi
3 hours ago





Thanks, just updated the code, assuming you can have new properties as well.

– loretoparisi
3 hours ago













what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

– AZ_
3 hours ago





what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]

– AZ_
3 hours ago













@loretoparisi updated

– adiga
3 hours ago





@loretoparisi updated

– adiga
3 hours ago













@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

– AZ_
3 hours ago







@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.

– AZ_
3 hours ago















@AZ_ updated. Not sure if it will fail for any scenario

– adiga
3 hours ago





@AZ_ updated. Not sure if it will fail for any scenario

– adiga
3 hours ago











1














A recursive example,






var obj1 = {

        prop1: 1,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "bar"

        },

        prop7: {},

    }



    var obj2 = {

        prop1: 3,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "foobar"

        },

        prop6: "new",

        prop7: {},

        prop8: {},

    }



    var result = {};



    function compare(obj1, obj2, obj_) {

        for (let k in obj1) {

            var type = typeof obj1[k];

            if (type === 'object') {

                obj_[k] = {};

                if (!obj2[k]){

                    obj_[k] = false;

                }else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object)) {

                    obj_[k] = true;

                } else {

                    compare(obj1[k], obj2[k], obj_[k]);

                }

            } else {

                obj_[k] = (obj1[k] === obj2[k]);

            }



        }

    }



    if (Object.keys(obj1).length < Object.keys(obj2).length) { //check if both objects vary in length.

        var tmp = obj1;

        obj1 = obj2;

        obj2 = tmp;

    }



    compare(obj1, obj2, result);



    console.log(result);








share|improve this answer


























  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a {} in the results for properties of type object that belong to both objects and that are {}.

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    58 mins ago
















1














A recursive example,






var obj1 = {

        prop1: 1,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "bar"

        },

        prop7: {},

    }



    var obj2 = {

        prop1: 3,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "foobar"

        },

        prop6: "new",

        prop7: {},

        prop8: {},

    }



    var result = {};



    function compare(obj1, obj2, obj_) {

        for (let k in obj1) {

            var type = typeof obj1[k];

            if (type === 'object') {

                obj_[k] = {};

                if (!obj2[k]){

                    obj_[k] = false;

                }else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object)) {

                    obj_[k] = true;

                } else {

                    compare(obj1[k], obj2[k], obj_[k]);

                }

            } else {

                obj_[k] = (obj1[k] === obj2[k]);

            }



        }

    }



    if (Object.keys(obj1).length < Object.keys(obj2).length) { //check if both objects vary in length.

        var tmp = obj1;

        obj1 = obj2;

        obj2 = tmp;

    }



    compare(obj1, obj2, result);



    console.log(result);








share|improve this answer


























  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a {} in the results for properties of type object that belong to both objects and that are {}.

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    58 mins ago














1












1








1







A recursive example,






var obj1 = {

        prop1: 1,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "bar"

        },

        prop7: {},

    }



    var obj2 = {

        prop1: 3,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "foobar"

        },

        prop6: "new",

        prop7: {},

        prop8: {},

    }



    var result = {};



    function compare(obj1, obj2, obj_) {

        for (let k in obj1) {

            var type = typeof obj1[k];

            if (type === 'object') {

                obj_[k] = {};

                if (!obj2[k]){

                    obj_[k] = false;

                }else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object)) {

                    obj_[k] = true;

                } else {

                    compare(obj1[k], obj2[k], obj_[k]);

                }

            } else {

                obj_[k] = (obj1[k] === obj2[k]);

            }



        }

    }



    if (Object.keys(obj1).length < Object.keys(obj2).length) { //check if both objects vary in length.

        var tmp = obj1;

        obj1 = obj2;

        obj2 = tmp;

    }



    compare(obj1, obj2, result);



    console.log(result);








share|improve this answer















A recursive example,






var obj1 = {

        prop1: 1,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "bar"

        },

        prop7: {},

    }



    var obj2 = {

        prop1: 3,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "foobar"

        },

        prop6: "new",

        prop7: {},

        prop8: {},

    }



    var result = {};



    function compare(obj1, obj2, obj_) {

        for (let k in obj1) {

            var type = typeof obj1[k];

            if (type === 'object') {

                obj_[k] = {};

                if (!obj2[k]){

                    obj_[k] = false;

                }else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object)) {

                    obj_[k] = true;

                } else {

                    compare(obj1[k], obj2[k], obj_[k]);

                }

            } else {

                obj_[k] = (obj1[k] === obj2[k]);

            }



        }

    }



    if (Object.keys(obj1).length < Object.keys(obj2).length) { //check if both objects vary in length.

        var tmp = obj1;

        obj1 = obj2;

        obj2 = tmp;

    }



    compare(obj1, obj2, result);



    console.log(result);








var obj1 = {

        prop1: 1,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "bar"

        },

        prop7: {},

    }



    var obj2 = {

        prop1: 3,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "foobar"

        },

        prop6: "new",

        prop7: {},

        prop8: {},

    }



    var result = {};



    function compare(obj1, obj2, obj_) {

        for (let k in obj1) {

            var type = typeof obj1[k];

            if (type === 'object') {

                obj_[k] = {};

                if (!obj2[k]){

                    obj_[k] = false;

                }else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object)) {

                    obj_[k] = true;

                } else {

                    compare(obj1[k], obj2[k], obj_[k]);

                }

            } else {

                obj_[k] = (obj1[k] === obj2[k]);

            }



        }

    }



    if (Object.keys(obj1).length < Object.keys(obj2).length) { //check if both objects vary in length.

        var tmp = obj1;

        obj1 = obj2;

        obj2 = tmp;

    }



    compare(obj1, obj2, result);



    console.log(result);





var obj1 = {

        prop1: 1,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "bar"

        },

        prop7: {},

    }



    var obj2 = {

        prop1: 3,

        prop2: "foo",

        prop3: {

            prop4: 2,

            prop5: "foobar"

        },

        prop6: "new",

        prop7: {},

        prop8: {},

    }



    var result = {};



    function compare(obj1, obj2, obj_) {

        for (let k in obj1) {

            var type = typeof obj1[k];

            if (type === 'object') {

                obj_[k] = {};

                if (!obj2[k]){

                    obj_[k] = false;

                }else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object)) {

                    obj_[k] = true;

                } else {

                    compare(obj1[k], obj2[k], obj_[k]);

                }

            } else {

                obj_[k] = (obj1[k] === obj2[k]);

            }



        }

    }



    if (Object.keys(obj1).length < Object.keys(obj2).length) { //check if both objects vary in length.

        var tmp = obj1;

        obj1 = obj2;

        obj2 = tmp;

    }



    compare(obj1, obj2, result);



    console.log(result);






share|improve this answer














share|improve this answer



share|improve this answer








edited 59 mins ago

























answered 1 hour ago









Shoyeb SheikhShoyeb Sheikh

652211




652211













  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a {} in the results for properties of type object that belong to both objects and that are {}.

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    58 mins ago



















  • Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a {} in the results for properties of type object that belong to both objects and that are {}.

    – loretoparisi
    1 hour ago








  • 1





    @loretoparisi I made a change, try now.

    – Shoyeb Sheikh
    58 mins ago

















Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a {} in the results for properties of type object that belong to both objects and that are {}.

– loretoparisi
1 hour ago







Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a {} in the results for properties of type object that belong to both objects and that are {}.

– loretoparisi
1 hour ago






1




1





@loretoparisi I made a change, try now.

– Shoyeb Sheikh
58 mins ago





@loretoparisi I made a change, try now.

– Shoyeb Sheikh
58 mins ago


















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