Solved

SQL Group by with Summing in Sub-select

Posted on 2009-06-29
4
753 Views
Last Modified: 2012-05-07
Hi All,

Quite new to the SQL Language and have been scratching my head with a group by with sub-select's summing some costs from another couple of tables.

Here is the statement;

SELECT     Account_Id AS Expr1,

(SELECT SUM(TOTAL_COST) AS Expr2 FROM Job_Labour WHERE Job_Labour.JOB_NUMBER = Job_Unfinished.Job_Number) AS Labour,
(SELECT SUM(Amount) AS Expr3 FROM Creditor_Invoice_Expense WHERE (Creditor_Invoice_Expense.Job_Number = Job_Unfinished.Job_Number)) AS Material_Cost

FROM  Job_Unfinished

WHERE (Major_Work_In_Progress = 0)

GROUP BY Account_Id, Job_Number

Which gives me the result of;

65      825.92      1240.45
652      86.10      NULL
652      49.08      NULL
652      NULL      NULL
652      38.34      NULL

What I am trying to do is group the Account_ID's together as one total.

ie.

65      825.92      1240.45
652      173.52      NULL

 If I omit 'Job_Number' from the GROUP BY statement it bemoans about 'Column 'Job_Unfinished.Job_Number' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.' of course.

Anybody help me out or point me in the right direction.

With thanks, Mark

0
Comment
Question by:ozbailer
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
4 Comments
 
LVL 60

Expert Comment

by:Kevin Cross
ID: 24742319
You can make your query a derived table and then sum on that grouped by just the accountid or rewrite query using left join instead of sub queries.
0
 
LVL 60

Accepted Solution

by:
Kevin Cross earned 500 total points
ID: 24742325
Using what you have as a derived table:
SELECT Account_Id
, SUM(Labour) AS Labour
, SUM(Material_Cost) AS Material_Cost
FROM (SELECT     Account_Id, 
 
(SELECT SUM(TOTAL_COST) AS Expr2 FROM Job_Labour WHERE Job_Labour.JOB_NUMBER = Job_Unfinished.Job_Number) AS Labour,
(SELECT SUM(Amount) AS Expr3 FROM Creditor_Invoice_Expense WHERE (Creditor_Invoice_Expense.Job_Number = Job_Unfinished.Job_Number)) AS Material_Cost
 
FROM  Job_Unfinished
 
WHERE (Major_Work_In_Progress = 0)
 
GROUP BY Account_Id, Job_Number
) t
GROUP BY Account_Id

Open in new window

0
 
LVL 60

Expert Comment

by:Kevin Cross
ID: 24742336
As LEFT JOINs:
SELECT ju.Account_Id, 
, SUM(jl.TOTAL_COST) AS Labour
, SUM(cie.Amount) AS Material_Cost
FROM Job_Unfinished ju
LEFT JOIN Job_Labour jl ON jl.JOB_NUMBER = ju.Job_Number
LEFT JOIN Creditor_Invoice_Expense cie ON cie.Job_Number = ju.Job_Number
WHERE (ju.Major_Work_In_Progress = 0)
GROUP BY Account_Id

Open in new window

0
 

Author Closing Comment

by:ozbailer
ID: 31598225
Cheers, Thank you!
0

Featured Post

Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Confronted with some SQL you don't know can be a daunting task. It can be even more daunting if that SQL carries some of the old secret codes used in the Ye Olde query syntax, such as: (+)     as used in Oracle;     *=     =*    as used in Sybase …
If you have heard of RFC822 date formats, they can be quite a challenge in SQL Server. RFC822 is an Internet standard format for email message headers, including all dates within those headers. The RFC822 protocols are available in detail at:   ht…
In a recent question (https://www.experts-exchange.com/questions/29004105/Run-AutoHotkey-script-directly-from-Notepad.html) here at Experts Exchange, a member asked how to run an AutoHotkey script (.AHK) directly from Notepad++ (aka NPP). This video…

751 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question