Escape Double Quote

  • Good morning. I am trying to find all instances of a string that contain the letters FSC.

    While I am able to find most of them, I am unable to find the ones wrapped in double quotes.

    Query example:

    Select *

    from myTable

    Where item like '%FSC%'

    This works great with the exception of when FSC is surrounded by double quotes.

    ex: MySentencecontains"FSC"

    I have tried Replace with no luck.

    Thanks for any assistance.

  • mbrady5 (9/14/2015)


    Good morning. I am trying to find all instances of a string that contain the letters FSC.

    While I am able to find most of them, I am unable to find the ones wrapped in double quotes.

    Query example:

    Select *

    from myTable

    Where item like '%FSC%'

    This works great with the exception of when FSC is surrounded by double quotes.

    ex: MySentencecontains"FSC"

    I have tried Replace with no luck.

    Thanks for any assistance.

    I suspect there is something else going on here because as posted this will work fine. Of course the leading wildcard is a performance problem but that is another topic.

    if OBJECT_ID('tempdb..#Something') is not null

    drop table #Something

    create table #Something

    (

    MySentence varchar(25)

    )

    insert #Something

    select '"FSC"'

    select *

    from #Something

    where MySentence like '%FSC%'

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  • I suspect there is a piece of info missing. Are you sure they are actually stored as double quotes.

    For example try this:

    DECLARE @Test TABLE(myString VARCHAR(25))

    INSERT INTO @Test

    VALUES ('TEST FSC'), ('MySentencecontains"FSC"')

    SELECT * FROM @Test

    SELECT * FROM @Test WHERE myString LIKE '%FSC%'

    Both statement returns both values including the one with double quotes.


    SELECT quote FROM brain WHERE original = 1
    0 rows returned

  • Tried with both the options below. It seems to be working fine.:-)

    SELECT * FROM

    (

    SELECT '"DFDJFDJLKLK"FSC"' COL1

    )a

    WHERE COL1 like '%FSC%'

    ----using replace

    SELECT * FROM

    (

    SELECT '"DFDJFDJLKLK"FSC"' COL1

    )a

    WHERE REPLACE(COL1,'"','') like '%FSC%'

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