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MSSQL Select Sum where Sum>X Problem

Posted on 2012-12-20
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Last Modified: 2012-12-20
Will someone please show me how to structure the following SQL statement so that it will work?  I do not know how to write it so that it pulls data where the sum of sales is greater than X.

SELECT crm_Customer_Employee_Assignments.CustomerNum, crm_customerPurchaseRollup.custLevel, sum(custsls.priceextension) as sales  
FROM crm_Customer_Employee_Assignments
LEFT OUTER JOIN CUSTSLS ON crm_Customer_Employee_Assignments.CustomerNum=CUSTSLS.CustNum
LEFT OUTER JOIN crm_customerPurchaseRollup ON crm_Customer_Employee_Assignments.CustomerNum = crm_customerPurchaseRollup.trueCustNum
WHERE SUM(CUSTSLS.PRICEEXTENSION)>=1.23 group by customernum
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Question by:trbbhm
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10 Comments
 
LVL 32

Expert Comment

by:Daniel Wilson
ID: 38709790
for aggregate conditions use HAVING, not WHERE.

Try this:

SELECT crm_Customer_Employee_Assignments.CustomerNum, crm_customerPurchaseRollup.custLevel, sum(custsls.priceextension) as sales  
FROM crm_Customer_Employee_Assignments
LEFT OUTER JOIN CUSTSLS ON crm_Customer_Employee_Assignments.CustomerNum=CUSTSLS.CustNum
LEFT OUTER JOIN crm_customerPurchaseRollup ON crm_Customer_Employee_Assignments.CustomerNum = crm_customerPurchaseRollup.trueCustNum
HAVING SUM(CUSTSLS.PRICEEXTENSION)>=1.23 group by customernum
0
 
LVL 10

Expert Comment

by:ienaxxx
ID: 38709791
What about

SELECT crm_Customer_Employee_Assignments.CustomerNum, crm_customerPurchaseRollup.custLevel, sum(custsls.priceextension) as sales  
FROM crm_Customer_Employee_Assignments
LEFT OUTER JOIN CUSTSLS ON crm_Customer_Employee_Assignments.CustomerNum=CUSTSLS.CustNum
LEFT OUTER JOIN crm_customerPurchaseRollup ON crm_Customer_Employee_Assignments.CustomerNum = crm_customerPurchaseRollup.trueCustNum
WHERE sales>=1.23 group by customernum 

Open in new window

0
 
LVL 11

Expert Comment

by:David Kroll
ID: 38709793
Do you mean you want to be able to use a variable called x?

DECLARE x double

SET x = 1.23

SELECT crm_Customer_Employee_Assignments.CustomerNum, crm_customerPurchaseRollup.custLevel, sum(custsls.priceextension) as sales  
FROM crm_Customer_Employee_Assignments
LEFT OUTER JOIN CUSTSLS ON crm_Customer_Employee_Assignments.CustomerNum=CUSTSLS.CustNum
LEFT OUTER JOIN crm_customerPurchaseRollup ON crm_Customer_Employee_Assignments.CustomerNum = crm_customerPurchaseRollup.trueCustNum
WHERE SUM(CUSTSLS.PRICEEXTENSION)>= x group by customernum
0
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Author Comment

by:trbbhm
ID: 38709812
@IENAXXX:  Invalid column name 'sales'.

@DKROLLCTN:  I was just using X as an example.  Say I wanted to pull data where sum(priceextension)> 450.64.  Your SQL example yielded the following error:  An aggregate may not appear in the WHERE clause unless it is in a subquery contained in a HAVING clause or a select list, and the column being aggregated is an outer reference.
0
 

Author Comment

by:trbbhm
ID: 38709824
@DANIELWILSON:  Incorrect syntax near the keyword 'group'.
0
 
LVL 32

Accepted Solution

by:
Daniel Wilson earned 500 total points
ID: 38709831
Adjusting mine a little ...

SELECT crm_Customer_Employee_Assignments.CustomerNum, crm_customerPurchaseRollup.custLevel, sum(custsls.priceextension) as sales  
FROM crm_Customer_Employee_Assignments
LEFT OUTER JOIN CUSTSLS ON crm_Customer_Employee_Assignments.CustomerNum=CUSTSLS.CustNum
LEFT OUTER JOIN crm_customerPurchaseRollup ON crm_Customer_Employee_Assignments.CustomerNum = crm_customerPurchaseRollup.trueCustNum
 group by customernum ,crm_customerPurchaseRollup.custLevel
HAVING SUM(CUSTSLS.PRICEEXTENSION)>=1.23
0
 
LVL 11

Expert Comment

by:David Kroll
ID: 38709838
SELECT crm_Customer_Employee_Assignments.CustomerNum, crm_customerPurchaseRollup.custLevel, sum(custsls.priceextension) as sales  
FROM crm_Customer_Employee_Assignments
LEFT OUTER JOIN CUSTSLS ON crm_Customer_Employee_Assignments.CustomerNum=CUSTSLS.CustNum
LEFT OUTER JOIN crm_customerPurchaseRollup ON crm_Customer_Employee_Assignments.CustomerNum = crm_customerPurchaseRollup.trueCustNum
group by customernum
HAVING SUM(CUSTSLS.PRICEEXTENSION)>= x
0
 

Author Closing Comment

by:trbbhm
ID: 38709843
CHA-CHING!!!  Thank you!!!
0
 
LVL 40

Expert Comment

by:Kyle Abrahams
ID: 38709895
Edit:  Type after user answered question.
0
 

Author Comment

by:trbbhm
ID: 38709922
Who's Dan?  Is this DKrollCTN?  Is this about the points that were awarded?

DanielWilson posted a suggestion, which I tested, and accepted as solution before I saw the post by DKrollCTN (which is virtually identical to the post by DanielWilson).

I really don't mind sharing points, but the *first* solution was accepted.  If there is an argument about who gets the points, please take it up with management.
0

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