Re: syntax error in plperl wrapper for Perl regex replace.

Le 25/08/18 à 01:55, io.sys a écrit :

Not being a perl expert either I think
the problem lies in the regex line.

You should eval it to have the expected
result (and you can use your 4th parameter again then).

eval "s/$_[1]/$_[2]/$_[3]";

Regards, Cédric

Tom,

You are right, it is a Perl's fault, not Postgre's.

You are right also with assumption, that the "s" command of Perl does not accept variable
in every of it's part. I have narrowed down the problem (btw, I could not find anything
about it is Perl docs) to the options part, ie. after last slash. Perl only accepts literals
there, not a variable of any kind. So these options must be hardcoded, not a dynamic value
constructed at runtime.

This compiles:
s/$_[1]/$_[2]/ig;

But even then, the command does not work as expected, as it does not interpret the
backreferences in a replacement string passed to it. It uses it literally. For example,
one would expect that the following call will return 'Praha':

select perl_regexp_replace ('PRAHA', '^(.)(.*)$', '$1\L$2');

But it returns:
$1\L$2

Well, Perl is praised for it's capabilities at the field of text processing, but this
experience disappointed me a little bit. Never mind, I will solve the things different way.

Thank you all for your inputs.

Regards, Roman Golis.