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Selecting specific characters in a field

  • May 15, 2011 at 9:25 am

    #73024

    How can I get the 8 characters that follow the characters 9^ in a field?

  • May 15, 2011 at 10:34 am

    #1324716

    Is this what you are looking fo:

    DECLARE @T VARCHAR(12)

    SET @T = '9^1234567890'

    SELECT CHARINDEX ( '^' ,@T,1) -- just so you understand what this performs

    SELECT SUBSTRING(@T,CHARINDEX ( '^' ,@T,1)+1 , 8)

    Result: 12345678

    If everything seems to be going well, you have obviously overlooked something.

    Ron

    Please help us, help you -before posting a question please read[/url]
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  • May 15, 2011 at 3:52 pm

    #1324790

    Just in case you're looking for a solution to recognize both characters (code snippet heavily borrowed from Ron):

    DECLARE @T VARCHAR(12)

    SET @T = 'xy9^1234567890'

    SELECT PATINDEX ( '%9^%' ,@T)

    SELECT SUBSTRING(@T,PATINDEX ( '%9^%' ,@T)+2 , 8)


    Lutz
    A pessimist is an optimist with experience.

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