31

I am not getting my head around this, and wondered if anyone may be able to help me with this.

I have 2 Tables called RES_DATA and INV_DATA

RES_DATA Contains my Customer as below

CUSTOMER ID | NAME

1, Robert
2, John
3, Peter

INV_DATA Contains their INVOICES as Below

INVOICE ID | CUSTOMER ID | AMOUNT

100, 1, £49.95
200, 1, £105.95
300, 2, £400.00
400, 3, £150.00
500, 1, £25.00

I am Trying to write a SELECT STATEMENT Which will give me the results as Below.

CUSTOMER ID | NAME | TOTAL AMOUNT

1, Robert, £180.90
2, John, £400.00
3, Peter, £150.00

I think I need 2 INNER JOINS Somehow to Add the tables and SUM Values of the INVOICES Table GROUPED BY the Customer Table but honestly think I am missing something. Can't even get close to the Results I need.

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1
  • what kind of SQL? SQL Server? Commented Apr 12, 2014 at 13:44

5 Answers 5

Reset to default
44

This should work.

SELECT a.[CUSTOMER ID], a.[NAME], SUM(b.[AMOUNT]) AS [TOTAL AMOUNT]
FROM RES_DATA a INNER JOIN INV_DATA b
ON a.[CUSTOMER ID]=b.[CUSTOMER ID]
GROUP BY a.[CUSTOMER ID], a.[NAME]

I tested it with SQL Fiddle against SQL Server 2008: http://sqlfiddle.com/#!3/1cad5/1

Basically what's happening here is that, because of the join, you are getting the same row on the "left" (i.e. from the RES_DATA table) for every row on the "right" (i.e. the INV_DATA table) that has the same [CUSTOMER ID] value. When you group by just the columns on the left side, and then do a sum of just the [AMOUNT] column from the right side, it keeps the one row intact from the left side, and sums up the matching values from the right side.

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2 Comments

The reason why a.[NAME] is also included is tuples which have those attributes are repeated. Am I right?
How would one optimize the solution, if performance is a factor?
10

Two ways to do it...

GROUP BY

SELECT RES.[CUSTOMER ID], RES,NAME, SUM(INV.AMOUNT) AS [TOTAL AMOUNT]
FROM RES_DATA RES
JOIN INV_DATA INV ON RES.[CUSTOMER ID] INV.[CUSTOMER ID]
GROUP BY RES.[CUSTOMER ID], RES,NAME

OVER

SELECT RES.[CUSTOMER ID], RES,NAME, 
       SUM(INV.AMOUNT) OVER (PARTITION RES.[CUSTOMER ID]) AS [TOTAL AMOUNT]
FROM RES_DATA RES
JOIN INV_DATA INV ON RES.[CUSTOMER ID] INV.[CUSTOMER ID]
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3

Use subquery

SELECT * FROM RES_DATA inner join (SELECT [CUSTOMER ID], sum([TOTAL AMOUNT]) FROM INV_DATA group by [CUSTOMER ID]) T on RES_DATA.[CUSTOMER ID] = t.[CUSTOMER ID]

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4 Comments

This answer will have O(n^2) performance, as op. to a group by or over which will have O(n)
Indeed, but he was looking for a result, not for performance.
This answer works well when you join within the same table to get a group by within some column in the table for each line in the table.
This solution has serious performance issues. Please do not consider this solution.
0

If you need to retrieve more columns other than columns which are in group by then you can consider below query. Check it once whether it is performing well or not. 

SELECT 
a.[CUSTOMER ID], 
a.[NAME], 
(select SUM(b.[AMOUNT]) from INV_DATA b
where b.[CUSTOMER ID] = a.[CUSTOMER ID]
GROUP BY b.[CUSTOMER ID]) AS [TOTAL AMOUNT]
FROM RES_DATA a
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Comments

-2 
SELECT RES.[CUSTOMER ID], RES.NAME, SUM(INV.AMOUNT) AS [TOTAL AMOUNT]
FROM RES_DATA
JOIN INV_DATA ON RES.[CUSTOMER ID]=INV.[CUSTOMER ID]
GROUP BY RES.[CUSTOMER ID], RES.NAME
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1 Comment

Please explain your code (even code comments are helpful).

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