SQL Query with multiple JOINS and junction table
I'm working with MySQL on this query and struggling a bit. The joins make sense to me but getting the data out of the images table which has a junction table called user_image seems difficult and I just can't grasp it.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) *
cos( radians( lat ) ) *
cos( radians( lng ) - radians(-80.6628) ) +
sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance,
t.id, t.name, t.price, t.duration, d.description, u.fname,
i.image_path, i.image_name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
-- Need to get all images that match trip and is_main = 1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
My image tables look like this...
user_image: trip_id | image_id
image: id | image_name | is_active | is_main
Not sure if I'm supposed to be using another join, a union, a query in a query? Really at a loss, would appreciate some help :)
mysql
edited Aug 20 '18 at 2:18
Rick James
42.9k22259
asked Aug 2 '18 at 11:13
BryanBryan
1
bumped to the homepage by Community♦ 3 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I'm working with MySQL on this query and struggling a bit. The joins make sense to me but getting the data out of the images table which has a junction table called user_image seems difficult and I just can't grasp it.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) *
cos( radians( lat ) ) *
cos( radians( lng ) - radians(-80.6628) ) +
sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance,
t.id, t.name, t.price, t.duration, d.description, u.fname,
i.image_path, i.image_name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
-- Need to get all images that match trip and is_main = 1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
My image tables look like this...
user_image: trip_id | image_id
image: id | image_name | is_active | is_main
Not sure if I'm supposed to be using another join, a union, a query in a query? Really at a loss, would appreciate some help :)
mysql
edited Aug 20 '18 at 2:18
Rick James
42.9k22259
asked Aug 2 '18 at 11:13
BryanBryan
1
bumped to the homepage by Community♦ 3 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
The geo portion of the query was from an example by Google, they used 'having' in their example so that's why it's there. Does that make a difference for trying to get data out of the image table?
– Bryan
Aug 2 '18 at 11:50
add a comment |
I'm working with MySQL on this query and struggling a bit. The joins make sense to me but getting the data out of the images table which has a junction table called user_image seems difficult and I just can't grasp it.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) *
cos( radians( lat ) ) *
cos( radians( lng ) - radians(-80.6628) ) +
sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance,
t.id, t.name, t.price, t.duration, d.description, u.fname,
i.image_path, i.image_name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
-- Need to get all images that match trip and is_main = 1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
My image tables look like this...
user_image: trip_id | image_id
image: id | image_name | is_active | is_main
Not sure if I'm supposed to be using another join, a union, a query in a query? Really at a loss, would appreciate some help :)
mysql
edited Aug 20 '18 at 2:18
Rick James
42.9k22259
asked Aug 2 '18 at 11:13
BryanBryan
1
I'm working with MySQL on this query and struggling a bit. The joins make sense to me but getting the data out of the images table which has a junction table called user_image seems difficult and I just can't grasp it.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) *
cos( radians( lat ) ) *
cos( radians( lng ) - radians(-80.6628) ) +
sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance,
t.id, t.name, t.price, t.duration, d.description, u.fname,
i.image_path, i.image_name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
-- Need to get all images that match trip and is_main = 1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
My image tables look like this...
user_image: trip_id | image_id
image: id | image_name | is_active | is_main
Not sure if I'm supposed to be using another join, a union, a query in a query? Really at a loss, would appreciate some help :)
mysql
mysql
edited Aug 20 '18 at 2:18
Rick James
42.9k22259
asked Aug 2 '18 at 11:13
BryanBryan
1
edited Aug 20 '18 at 2:18
Rick James
42.9k22259
asked Aug 2 '18 at 11:13
BryanBryan
1
edited Aug 20 '18 at 2:18
Rick James
42.9k22259
edited Aug 20 '18 at 2:18
Rick James
42.9k22259
edited Aug 20 '18 at 2:18
Rick James
42.9k22259
42.9k22259
asked Aug 2 '18 at 11:13
BryanBryan
1
asked Aug 2 '18 at 11:13
BryanBryan
1
asked Aug 2 '18 at 11:13
BryanBryan
1
1
bumped to the homepage by Community♦ 3 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 3 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
The geo portion of the query was from an example by Google, they used 'having' in their example so that's why it's there. Does that make a difference for trying to get data out of the image table?
– Bryan
Aug 2 '18 at 11:50
add a comment |
The geo portion of the query was from an example by Google, they used 'having' in their example so that's why it's there. Does that make a difference for trying to get data out of the image table?
– Bryan
Aug 2 '18 at 11:50
The geo portion of the query was from an example by Google, they used 'having' in their example so that's why it's there. Does that make a difference for trying to get data out of the image table?
– Bryan
Aug 2 '18 at 11:50
The geo portion of the query was from an example by Google, they used 'having' in their example so that's why it's there. Does that make a difference for trying to get data out of the image table?
– Bryan
Aug 2 '18 at 11:50
add a comment |
1 Answer
1
active
oldest
votes
Ok I think I got it. Here's the working example. Thanks to everyone that responded.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) * cos( radians( lat ) )
* cos( radians( lng ) - radians(-80.6628) ) + sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance, t.id, t.name, t.price, t.duration, d.description, u.fname, i.path, i.name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
LEFT OUTER JOIN user_image ui ON ui.trip_id = t.id
LEFT OUTER JOIN image i ON ui.image_id = i.id
AND i.main=1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
answered Aug 2 '18 at 13:51
BryanBryan
1
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "182"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdba.stackexchange.com%2fquestions%2f213888%2fsql-query-with-multiple-joins-and-junction-table%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Ok I think I got it. Here's the working example. Thanks to everyone that responded.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) * cos( radians( lat ) )
* cos( radians( lng ) - radians(-80.6628) ) + sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance, t.id, t.name, t.price, t.duration, d.description, u.fname, i.path, i.name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
LEFT OUTER JOIN user_image ui ON ui.trip_id = t.id
LEFT OUTER JOIN image i ON ui.image_id = i.id
AND i.main=1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
answered Aug 2 '18 at 13:51
BryanBryan
1
add a comment |
Ok I think I got it. Here's the working example. Thanks to everyone that responded.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) * cos( radians( lat ) )
* cos( radians( lng ) - radians(-80.6628) ) + sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance, t.id, t.name, t.price, t.duration, d.description, u.fname, i.path, i.name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
LEFT OUTER JOIN user_image ui ON ui.trip_id = t.id
LEFT OUTER JOIN image i ON ui.image_id = i.id
AND i.main=1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
answered Aug 2 '18 at 13:51
BryanBryan
1
add a comment |
Ok I think I got it. Here's the working example. Thanks to everyone that responded.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) * cos( radians( lat ) )
* cos( radians( lng ) - radians(-80.6628) ) + sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance, t.id, t.name, t.price, t.duration, d.description, u.fname, i.path, i.name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
LEFT OUTER JOIN user_image ui ON ui.trip_id = t.id
LEFT OUTER JOIN image i ON ui.image_id = i.id
AND i.main=1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
answered Aug 2 '18 at 13:51
BryanBryan
1
Ok I think I got it. Here's the working example. Thanks to everyone that responded.
SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) * cos( radians( lat ) )
* cos( radians( lng ) - radians(-80.6628) ) + sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance, t.id, t.name, t.price, t.duration, d.description, u.fname, i.path, i.name
FROM city c
JOIN trip t ON c.id = t.city_id
JOIN trip_description d ON t.id = d.trip_id
JOIN user u ON t.user_id = u.id
LEFT OUTER JOIN user_image ui ON ui.trip_id = t.id
LEFT OUTER JOIN image i ON ui.image_id = i.id
AND i.main=1
HAVING distance < 20
ORDER BY distance
LIMIT 0 , 45;
answered Aug 2 '18 at 13:51
BryanBryan
1
answered Aug 2 '18 at 13:51
BryanBryan
1
answered Aug 2 '18 at 13:51
BryanBryan
1
answered Aug 2 '18 at 13:51
BryanBryan
1
1
add a comment |
add a comment |
Thanks for contributing an answer to Database Administrators Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdba.stackexchange.com%2fquestions%2f213888%2fsql-query-with-multiple-joins-and-junction-table%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
The geo portion of the query was from an example by Google, they used 'having' in their example so that's why it's there. Does that make a difference for trying to get data out of the image table?
– Bryan
Aug 2 '18 at 11:50