Yes it does :o) Thank you! The methods that work (allowing for spaces) is the charindex method and this one.
Now the question is, which is more efficient? Using Stefan's carindex method or this one? This will be part of a maintenance routine and there will could be thousands of records that match.
>Would this work for you
>SELECT * FROM Mytable WHERE REPLACE(Mycolumn, SPACE(1), SPACE(0)) LIKE '%;ABC'
>
?
>
>>Perfect Stefan, thanks!
>>
.·*´¨)
.·`TCH
(..·*
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