Re: An Oracle 10g Question

On Jan 26, 4:01 pm, "S. Anthony Sequeira" <t..._at_sequeira.org.uk> wrote:
> On Sat, 2008-01-26 at 13:50 -0800, ame..._at_iwc.net wrote:
> > On Jan 25, 9:46 pm, csn <csn..._at_gmail.com> wrote:
> > > select
> > >     regexp_instr('"null","null","null","GMO, DXDT, SDS, QID,
> > > MZZ","90", "3 ASC"',
> > >     '"[[:alnum:]]+, .*"',1) pos from dual;
>
> > > This one looks for
>
> > > "XXXXXXX,_......."
>
> > > which is exactly your description, and returns a non-zero code of 22
> > > which is the starting position in the string.
>
> > > Note: _ means space.
>
> > Last question:  Your example seems to work.  Now, I'm trying to
> > understand it in english:
>
> > There is the string:  '"null","null","null","GMO, DXDT, SDS, QID,
> > MZZ","90", "3 ASC"',
> > And you are looking for one or more alphanumeric characters....
>
> > And then I get lost with the .*"',1)
>
> csn has already explained it "XXXXXXX,_......."
>
> Attempting to match:
> doublequoteoneormorealphanumericcharscommaspacethenzeroormoreofanycharacter­doublequote
>
> As I see it
> --
> S. Anthony Sequeira
> ++
> "Everyone's head is a cheap movie show."
> -- Jeff G. Bone
> ++- Hide quoted text -
>
> - Show quoted text -

Everyone's help is much appreciated. Regualr Expressions have always been difficult to understand, though I continue to try. Received on Sat Jan 26 2008 - 16:33:27 CST