something like:

select pmd.psp_item_no,pmd.psp_item_var,sum(psp_qty_prpnl) as Total

from pmddb..pmd_mpsp_ps_postn pmd

--== added freshly

...

calvo (6/9/2011)

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Took me a second to understand it too.

I believe the logic is, if there is one and only one entity for the RU, then show it. (adhering to your...

Try the MERGE command!

How abt this:

select CartesisRU

from #Cartesis_to_hfm

group by CartesisRU

HAVING MAX(HFMEntity) = MIN(HFMEntity)

How abt this?

; with allServicesRenderedCustomers As

(

select c.CustomerName

from #Customer c

group by

c.CustomerName

HAVING AVG ( CAST( c.Rendered as decimal(4,2))) = 1.00

)

select *

from #Customer c

join allServicesRenderedCustomers a

...

I dont get clearly what u mean by another table. can u describe more ? Sample table(s), some sample data and expected result, to start with??

An AND clause to the WHERE clause, may be??

Like, WHERE Description = 'Optional' ??

If the date will always an ascending order, then this will also help:

SELECT OuterTable.DateValues , OuterTable.IDs ,

[Effective Start] = OuterTable.DateValues ,

...

This MAY NOT be the best code or the best-performant one, but this cuts the deal.

Sample data ( for anyone who is interested in solving this problem)

SET NOCOUNT ON

--= This...

Look at UNPIVOT in Books Online ( a free help that comes with SQL Server). Online link to UNPIVOT - http://msdn.microsoft.com/en-us/library/ms177410.aspx

This problem is hard to solve unless we have another ID column which defines an unique row number per row; like an identity column. Without order of rows, the results...

Try this:

SELECT LEFT( WkstationID, 3 ) WkstationID , COUNT(*) COUNTS

FROM Workstation

GROUP BY LEFT( WkstationID, 3 )

try this:

DECLARE @Table TABLE

( Value CHAR(2))

INSERT INTO @Table VALUES

('1d'),

('1y'),

('2d'),

('3d'),

('2w'),

('3y'),

('2m'),

('3m')

SELECT Value

FROM @Table

ORDER BY RIGHT(Value,1) ,CAST ( LEFT(Value,1) AS INT)

{Edit: Edited the code as per the expected result}

Dynamic Cross Tabs!!

Read this article[/url] from Jeff Moden on how to acheive that!

Lowell (5/27/2011)

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ColdCoffee (5/27/2011)

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Lowell, actually, if u look at the new dataset the OP provided, there are dups.. so, a GROUP BY of managerID with COUNT > 1 wont work on...