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How can I select and then update certain characters of a string in sql server 2005?

Posted on 2014-03-17
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Last Modified: 2014-03-17
Hi

Suppose I have a column with data like

256dddggg@@576576
dgg782t@@6786
cb8979000shdj@@89789789
cn44@@123

or any other combination.

I need to first select this part   @@576576 and remove the @@ so that I am left with  576576. Then I want to select everything before @@ and update the column with whats left hence removing everything after @@.

Hope this makes sense.
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Question by:jazz__man
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4 Comments
 
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Assisted Solution

by:lcohan
lcohan earned 166 total points
ID: 39934934
You mean like this?

select LEFT('cb8979000shdj@@89789789', charindex('@@','cb8979000shdj@@89789789')-1)


If that's what you need just replace 'cb8979000shdj@@89789789' with any other string you posted above to test it then with the actual column name which you can use in an UPDATE statement like:

update tablename
set columnname = LEFT(columnname, charindex('@@',columnname)-1)
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LVL 38

Assisted Solution

by:Jim P.
Jim P. earned 167 total points
ID: 39934938
This should do it for you:

select	left(colName, CHARINDEX(CHAR(64),colName,1)-1) -- CHAR(64) = @
from	MyTable

Open in new window


or to update
Update  MyTable
set        colName = left(colName, CHARINDEX(CHAR(64),colName,1)-1) -- CHAR(64) = @
where  colName like '%' + CHAR(64) + '%'

Open in new window

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Accepted Solution

by:
Kyle Abrahams earned 167 total points
ID: 39934940
declare @v varchar(50)
set @v = 'dgg782t@@6786'

select substring(@v, 1, charindex('@',@v) - 1) LEFTOF,
substring(@v, charindex('@@',@v) + 2, len(@v) - charindex('@@',@v) + 2) RIGHTOF
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Expert Comment

by:lcohan
ID: 39934956
Also what do you want to do with the characters AFTER the @@? You say "I need to first select this part   @@576576 and remove the @@ so that I am left with  576576." but not what to do with those and here's how you can get that data out of any of the strings you posted above -or columnname:

select substring('cb8979000shdj@@89789789', charindex('@@','cb8979000shdj@@89789789')+2, LEN('cb8979000shdj@@89789789')-2)
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