Обсуждение: regexp replace?

I'd like to create a function that will take strings
like:

abcdef 'quote1' "quote2" / xyz etc.

and return:

abcdef-quote1-quote2-xyz-etc.

Basically remove all characters that aren't
[A-Za-z0-9], -, or ., and replace all spaces with
hyphens. Possible? The only string function that
supports regular expressions is substring and I
couldn't make that do what I want.

TIA,
CSN




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On Thu, 6 May 2004, CSN wrote:

> I'd like to create a function that will take strings
> like:
>
> abcdef 'quote1' "quote2" / xyz etc.
>
> and return:
>
> abcdef-quote1-quote2-xyz-etc.
>
> Basically remove all characters that aren't
> [A-Za-z0-9], -, or ., and replace all spaces with
> hyphens. Possible? The only string function that
> supports regular expressions is substring and I
> couldn't make that do what I want.

I think translate may be able to do what you want pretty handily.  May
take two passes, i.e. a subselect.

On Thu, May 06, 2004 at 01:59:16PM -0700, CSN wrote:

> Basically remove all characters that aren't
> [A-Za-z0-9], -, or ., and replace all spaces with
> hyphens. Possible? The only string function that
> supports regular expressions is substring and I
> couldn't make that do what I want.

This should be trivial with plperl ... I'd think you are wasting your
time trying to do it with plpgsql or SQL; this is what Postgres
flexibility is for!

--
Alvaro Herrera (<alvherre[a]dcc.uchile.cl>)
"Ni aun el genio muy grande llegaría muy lejos
si tuviera que sacarlo todo de su propio interior" (Goethe)