Combing two queries in one












0















I have two tables, e.g. "Locations" and "Connections"



"Locations" has values



Id | Dimension
---------------
1 | 4
2 | 8
3 | 2


"Connections" maintains attributes



Origin | Destination | Value | Distance_KM
-------------------------------------------
1 | 2 | 500 | 30
1 | 3 | 100 | 20
2 | 1 | 100 | 10
2 | 3 | 300 | 10
3 | 1 | 100 | 40


I want to create an output with the following Attribute Table. Where "In" correspond to "Destination" from "Connections" and "Out" to "Origin" accordingly.



Id | Dimension | In_Value | In_Count | In_Dist | Out_Value | Out_Count | Out_Dist
----------------------------------------------------------------------------------
1 | 4 | 200 | 2 | 50 | 600 | 2 | 50
2 | 8 | 500 | 1 | 30 | 400 | 2 | 20
3 | 2 | 400 | 2 | 30 | 100 | 1 | 40


I can achieve the result that I strive for separately with two queries.



Query 1



SELECT C.Destination, SUM(C.Value) AS In_Value, COUNT(C.Destination) AS In_Count, SUM(C.Distance_KM) AS In_Dist
FROM Connections AS C
GROUP BY C.Destination


Query 2



SELECT C.Origin, SUM(C.Value) AS Out_Value, COUNT(C.Origin) AS Out_Count, SUM(C.Origine_KM) AS Out_Dist
FROM Connections AS C
GROUP BY C.Origin


Nevertheless, there should be only one query that solves my issue, is not it? I tried this but no success.



SELECT L.Id AS Id, L.Dimension AS Dimension, C.In_Value, C.In_Count, C.In_Dist, C.Out_Value, C.Out_Count, C.Out_Dist
FROM Locations AS L
LEFT JOIN (
SELECT C.Destination, SUM(C.Value) AS In_Value, COUNT(C.Destination) AS In_Count, SUM(C.Distance_KM) AS In_Dist
FROM Connections AS C
GROUP BY C.Destination
) ON L.Id = C.Destination
LEFT JOIN (
SELECT C.Origin, SUM(C.Value) AS Out_Value, COUNT(C.Origin) AS Out_Count, SUM(C.Origine_KM) AS Out_Dist
FROM Connections AS C
GROUP BY C.Origin
) ON L.Id = C.Origin


Basically, I do not know if I eligible to add a second LEFT JOIN ON to the query with already existing LEFT JOIN ON, am I?





References:




  • Two SQL LEFT JOINS produce incorrect result

  • Multiple left joins on multiple tables in one query










share|improve this question









New contributor




Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • Specify your DBMS, including version.

    – Akina
    7 mins ago











  • I tried this but no success. Each subquery must have a separate alias unique within the whole query. ON clause must refer to a field with that alias, not a name iof inner table which is (in general) invisible out of the subquery.

    – Akina
    5 mins ago
















0















I have two tables, e.g. "Locations" and "Connections"



"Locations" has values



Id | Dimension
---------------
1 | 4
2 | 8
3 | 2


"Connections" maintains attributes



Origin | Destination | Value | Distance_KM
-------------------------------------------
1 | 2 | 500 | 30
1 | 3 | 100 | 20
2 | 1 | 100 | 10
2 | 3 | 300 | 10
3 | 1 | 100 | 40


I want to create an output with the following Attribute Table. Where "In" correspond to "Destination" from "Connections" and "Out" to "Origin" accordingly.



Id | Dimension | In_Value | In_Count | In_Dist | Out_Value | Out_Count | Out_Dist
----------------------------------------------------------------------------------
1 | 4 | 200 | 2 | 50 | 600 | 2 | 50
2 | 8 | 500 | 1 | 30 | 400 | 2 | 20
3 | 2 | 400 | 2 | 30 | 100 | 1 | 40


I can achieve the result that I strive for separately with two queries.



Query 1



SELECT C.Destination, SUM(C.Value) AS In_Value, COUNT(C.Destination) AS In_Count, SUM(C.Distance_KM) AS In_Dist
FROM Connections AS C
GROUP BY C.Destination


Query 2



SELECT C.Origin, SUM(C.Value) AS Out_Value, COUNT(C.Origin) AS Out_Count, SUM(C.Origine_KM) AS Out_Dist
FROM Connections AS C
GROUP BY C.Origin


Nevertheless, there should be only one query that solves my issue, is not it? I tried this but no success.



SELECT L.Id AS Id, L.Dimension AS Dimension, C.In_Value, C.In_Count, C.In_Dist, C.Out_Value, C.Out_Count, C.Out_Dist
FROM Locations AS L
LEFT JOIN (
SELECT C.Destination, SUM(C.Value) AS In_Value, COUNT(C.Destination) AS In_Count, SUM(C.Distance_KM) AS In_Dist
FROM Connections AS C
GROUP BY C.Destination
) ON L.Id = C.Destination
LEFT JOIN (
SELECT C.Origin, SUM(C.Value) AS Out_Value, COUNT(C.Origin) AS Out_Count, SUM(C.Origine_KM) AS Out_Dist
FROM Connections AS C
GROUP BY C.Origin
) ON L.Id = C.Origin


Basically, I do not know if I eligible to add a second LEFT JOIN ON to the query with already existing LEFT JOIN ON, am I?





References:




  • Two SQL LEFT JOINS produce incorrect result

  • Multiple left joins on multiple tables in one query










share|improve this question









New contributor




Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • Specify your DBMS, including version.

    – Akina
    7 mins ago











  • I tried this but no success. Each subquery must have a separate alias unique within the whole query. ON clause must refer to a field with that alias, not a name iof inner table which is (in general) invisible out of the subquery.

    – Akina
    5 mins ago














0












0








0








I have two tables, e.g. "Locations" and "Connections"



"Locations" has values



Id | Dimension
---------------
1 | 4
2 | 8
3 | 2


"Connections" maintains attributes



Origin | Destination | Value | Distance_KM
-------------------------------------------
1 | 2 | 500 | 30
1 | 3 | 100 | 20
2 | 1 | 100 | 10
2 | 3 | 300 | 10
3 | 1 | 100 | 40


I want to create an output with the following Attribute Table. Where "In" correspond to "Destination" from "Connections" and "Out" to "Origin" accordingly.



Id | Dimension | In_Value | In_Count | In_Dist | Out_Value | Out_Count | Out_Dist
----------------------------------------------------------------------------------
1 | 4 | 200 | 2 | 50 | 600 | 2 | 50
2 | 8 | 500 | 1 | 30 | 400 | 2 | 20
3 | 2 | 400 | 2 | 30 | 100 | 1 | 40


I can achieve the result that I strive for separately with two queries.



Query 1



SELECT C.Destination, SUM(C.Value) AS In_Value, COUNT(C.Destination) AS In_Count, SUM(C.Distance_KM) AS In_Dist
FROM Connections AS C
GROUP BY C.Destination


Query 2



SELECT C.Origin, SUM(C.Value) AS Out_Value, COUNT(C.Origin) AS Out_Count, SUM(C.Origine_KM) AS Out_Dist
FROM Connections AS C
GROUP BY C.Origin


Nevertheless, there should be only one query that solves my issue, is not it? I tried this but no success.



SELECT L.Id AS Id, L.Dimension AS Dimension, C.In_Value, C.In_Count, C.In_Dist, C.Out_Value, C.Out_Count, C.Out_Dist
FROM Locations AS L
LEFT JOIN (
SELECT C.Destination, SUM(C.Value) AS In_Value, COUNT(C.Destination) AS In_Count, SUM(C.Distance_KM) AS In_Dist
FROM Connections AS C
GROUP BY C.Destination
) ON L.Id = C.Destination
LEFT JOIN (
SELECT C.Origin, SUM(C.Value) AS Out_Value, COUNT(C.Origin) AS Out_Count, SUM(C.Origine_KM) AS Out_Dist
FROM Connections AS C
GROUP BY C.Origin
) ON L.Id = C.Origin


Basically, I do not know if I eligible to add a second LEFT JOIN ON to the query with already existing LEFT JOIN ON, am I?





References:




  • Two SQL LEFT JOINS produce incorrect result

  • Multiple left joins on multiple tables in one query










share|improve this question









New contributor




Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I have two tables, e.g. "Locations" and "Connections"



"Locations" has values



Id | Dimension
---------------
1 | 4
2 | 8
3 | 2


"Connections" maintains attributes



Origin | Destination | Value | Distance_KM
-------------------------------------------
1 | 2 | 500 | 30
1 | 3 | 100 | 20
2 | 1 | 100 | 10
2 | 3 | 300 | 10
3 | 1 | 100 | 40


I want to create an output with the following Attribute Table. Where "In" correspond to "Destination" from "Connections" and "Out" to "Origin" accordingly.



Id | Dimension | In_Value | In_Count | In_Dist | Out_Value | Out_Count | Out_Dist
----------------------------------------------------------------------------------
1 | 4 | 200 | 2 | 50 | 600 | 2 | 50
2 | 8 | 500 | 1 | 30 | 400 | 2 | 20
3 | 2 | 400 | 2 | 30 | 100 | 1 | 40


I can achieve the result that I strive for separately with two queries.



Query 1



SELECT C.Destination, SUM(C.Value) AS In_Value, COUNT(C.Destination) AS In_Count, SUM(C.Distance_KM) AS In_Dist
FROM Connections AS C
GROUP BY C.Destination


Query 2



SELECT C.Origin, SUM(C.Value) AS Out_Value, COUNT(C.Origin) AS Out_Count, SUM(C.Origine_KM) AS Out_Dist
FROM Connections AS C
GROUP BY C.Origin


Nevertheless, there should be only one query that solves my issue, is not it? I tried this but no success.



SELECT L.Id AS Id, L.Dimension AS Dimension, C.In_Value, C.In_Count, C.In_Dist, C.Out_Value, C.Out_Count, C.Out_Dist
FROM Locations AS L
LEFT JOIN (
SELECT C.Destination, SUM(C.Value) AS In_Value, COUNT(C.Destination) AS In_Count, SUM(C.Distance_KM) AS In_Dist
FROM Connections AS C
GROUP BY C.Destination
) ON L.Id = C.Destination
LEFT JOIN (
SELECT C.Origin, SUM(C.Value) AS Out_Value, COUNT(C.Origin) AS Out_Count, SUM(C.Origine_KM) AS Out_Dist
FROM Connections AS C
GROUP BY C.Origin
) ON L.Id = C.Origin


Basically, I do not know if I eligible to add a second LEFT JOIN ON to the query with already existing LEFT JOIN ON, am I?





References:




  • Two SQL LEFT JOINS produce incorrect result

  • Multiple left joins on multiple tables in one query







query aggregate table






share|improve this question









New contributor




Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 24 mins ago







Taras













New contributor




Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 29 mins ago









TarasTaras

1012




1012




New contributor




Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Taras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.













  • Specify your DBMS, including version.

    – Akina
    7 mins ago











  • I tried this but no success. Each subquery must have a separate alias unique within the whole query. ON clause must refer to a field with that alias, not a name iof inner table which is (in general) invisible out of the subquery.

    – Akina
    5 mins ago



















  • Specify your DBMS, including version.

    – Akina
    7 mins ago











  • I tried this but no success. Each subquery must have a separate alias unique within the whole query. ON clause must refer to a field with that alias, not a name iof inner table which is (in general) invisible out of the subquery.

    – Akina
    5 mins ago

















Specify your DBMS, including version.

– Akina
7 mins ago





Specify your DBMS, including version.

– Akina
7 mins ago













I tried this but no success. Each subquery must have a separate alias unique within the whole query. ON clause must refer to a field with that alias, not a name iof inner table which is (in general) invisible out of the subquery.

– Akina
5 mins ago





I tried this but no success. Each subquery must have a separate alias unique within the whole query. ON clause must refer to a field with that alias, not a name iof inner table which is (in general) invisible out of the subquery.

– Akina
5 mins ago










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