lee rite (3/3/2011)

---

lee rite (3/3/2011)

---

Thanks ColdCoffee! I think you nailed it. Your query selects all companies that never...

SELECT GUID

FROM #T

GROUP BY GUID

HAVING COUNT( DISTINCT CompanyName) = 1

lee rite (3/3/2011)

---

Craig Farrell (3/3/2011)

---

lee rite (3/3/2011)

---

Thanks for your help, guys! There are actually 25 columns, not 3.

Let me explain exactly what I'm working with, so you will see...

Try this:

SELECT ID

FROM

( SELECT * FROM #T ) PIVOT_TABLE

UNPIVOT

(

Vals FOR ColNames IN ([2008],[2009],[2010])

) PIVOT_HANDLE

GROUP BY ID

HAVING COUNT(...

Makes sense :-); how many rows are there in the table ?

This:

DECLARE @startDT DATETIME,

@endDT DATETIME

SELECT @startDT = 'Feb 1 2011 12:00AM',

...

NOthing, actualy 😛 two ways of writing it 🙂

Yes it will.. IF u could post sample data in a readily consumable foramt, we can work right away in providing a code for you..

Another one to the box:

select d.EmpNO ,

stuff( cast( (

select ','+ cast(...

Else:

SELECT distinct E.EmpNO,Y.Years As 'Missing Years' --E.EmpNo , Y.Years , d.EmpNO as 'DEmp', d.RYear

FROM demo E

CROSS JOIN

...

Try this:

; WITH DistinctEmpNos (EmpNo) AS

(

SELECT DISTINCT EmpNo

FROM demo

),

YearsAndEmps AS

(

SELECT E.EmpNo , Y.Years

FROM DistinctEmpNos E

...

riyaz.mohammed (3/1/2011)

---

ColdCoffee (3/1/2011)

---

Try this:

select *

from orders outer_table

where itemid = '1063'

and not exists ( select 1 from orders inner_table

...

Select * from table

where isnull ( (c1 +...

Craig Farrell (3/1/2011)

---

A question, how did you decide you only wanted 2 and 4? 1 also has a 1063, but has additional items, so in theory you could display...

Try this:

select *

from orders outer_table

where itemid = '1063'

and not exists ( select 1 from orders inner_table

...