Re: regexp_replace

How about:

select regexp_replace('71.09.6.01.3', '(\d)[.-]', '\1', 'g');

?

In your example, the (\d)[.-](\d) says find a digit followed by a period or
dash followed by another digit. The first time through 1.0 is matched and
replaced with 10 (710) with the "current location" pointing before the 9.
Go again and 9.6 is replaced by 96 for (71096) with the "current location"
pointing to the period! So ".0" doesn't match. (71096.0) next match is 1.3
and result is 13 ( 71096.013). If you don't want to eliminate the period or
dash unless it is _between_ two digits, try:

select regexp_replace('71.09.6.01.3', '(\d)[.-](?=\d)', '\1', 'g');

(?=\d) is a "look ahead") match which says that the period or dash must be
followed by a digit, but the expression _does not_ "consume" the digit
matched.

On Thu, Jan 14, 2016 at 1:43 PM, Andy Colson <andy(at)squeakycode(dot)net> wrote:

> Hi all.
>
> This is not doing as I'd expected:
>
> select regexp_replace('71.09.6.01.3', '(\d)[.-](\d)', '\1\2', 'g');
>
> regexp_replace
> ----------------
> 71096.013
> (1 row)
>
> It acts the same with dashes:
> select regexp_replace('71-09-6-01-3', '(\d)[.-](\d)', '\1\2', 'g');
>
> regexp_replace
> ----------------
> 71096-013
> (1 row)
>
> I cannot use translate because there is other text in the field. I'm
> trying to strip masking characters from a parcel number in a larger text
> field (for example: "the parcel 12-34-56 has caught on fire")
>
> I seem to be missing something, any hints?
>
> I'm on PG 9.3.9 on Slackware64.
>
> Thanks for your time,
>
> -Andy
>
>
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