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Thanks. At least now I can stop hunting for the way to do it. I have a book entitled "The Practical SQL Handbook" by Judith S. Bowman, et. al., Addison-Wesley 1996, which uses the syntax I list below in a LIKE statement, saying "Consult you system's documentation for the actual syntax." The book is based on something called Transact-SQL. Hmph.
Paul
Sybrand Bakker wrote:
> Oracle and SQL in general doesn't support regular expressions.
> You could trick this a bit by using
> replace(<column>,'0123456789','@@@@@@@@@@') like '@%'
> but that will definitely not use an index
>
> Hth,
>
> Sybrand Bakker, Oracle DBA
>
> "pretzelman" <pretzelman_at_fakeaddress.com> wrote in message
> news:B0A55195D3D4B294.3FBA51833427F821.162B86F9A16323EF_at_lp.airnews.net...
> > Hello,
> >
> > Does anyone know how Oracle allows you to specify a pattern such as
> > this:
> > LIKE '[0-9]%'
> > I am trying to get all values which begin with a numeric digit, followed
> > by any number of characters of anything else. Oracle doesn't like this;
> > IE it doesn't give me the rows I know are there. Thanks.
> >
> > Paul
> >
Received on Thu Jul 27 2000 - 00:00:00 CDT