Problem to apply replacement with regexp that include look ahead in Postgres

Question

I have this kind of string 12s2-3a abc def and I am trying to apply the regexp ^(?=.*\d) to remove the first part and having as left abc def.

I have tried with:

select substring ('12s2-3 abc def' from '^(?=.*\d)');

and

select regexp_replace ('12s2-3 abc def', '^(?=.*\d)', '');

but none of those has worked. The only one that looks working is the match verification (~).

Any idea?

Answer 1

Look-ahead are zero width. They don't match anything themselves. So your replacement operations is just replacing the empty string with the empty string, under certain conditions.

Simply removing the look-ahead designation makes it give the answer you want for your one example:

select regexp_replace ('12s2-3 abc def', '^.*\d', '');