This should work for you.
Declare @testtable table(name varchar(40))
Declare @string varchar(40)
Set @string = 'This is a test of the '''
Insert into @testtable (name)
Select @string
select *
from @testtable
where [name] like '%''%'
select replace([name],'''',''' Has been replaced')
From @testtable
Jason...AKA CirqueDeSQLeil
_______________________________________________
I have given a name to my pain...MCM SQL Server, MVP
SQL RNNR
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