How to replace in one regexp

appleii has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: How to replace in one regexp by Sue D. Nymme (Monk) on Sep 30, 2009 at 17:37 UTC
Written as two statements, your expressions are clear. Why confuse the issue by writing them as one? "Optimize" does not mean "use as few statements as possible."	[reply]
Re: How to replace in one regexp by Anonymous Monk on Sep 30, 2009 at 15:27 UTC
CGI, URI `#!/usr/bin/perl -- use strict; use warnings; use URI; my $path = { URI->new( "/index.php?prefix=/user/" )->query_form }->{pr +efix}; print "$path\n"; use CGI; $path = CGI->new( "prefix=/user/")->param("prefix"); print "$path\n"; __END__ /user/ /user/` [download]	[reply] [d/l]
Re^2: How to replace in one regexp by ikegami (Patriarch) on Sep 30, 2009 at 18:33 UTC
Good call to use URI, but your code does something different than the OP's. The following would be equivalent: `#!/bin/env perl use strict; use warnings; use URI qw( ); my $target = "/index.php?prefix=/user/"; my $path = URI->new_abs( '.', $target )->path(); print "$path\n";` [download]	[reply] [d/l]
Re: How to replace in one regexp by appleii (Novice) on Sep 30, 2009 at 16:51 UTC
`$path = substr($path, 0, rindex($path, "/", ($_ = rindex($path, "?")) +!= -1 ? $_ : length($path)) + 1);` [download]	[reply] [d/l]
Re: How to replace in one regexp by JavaFan (Canon) on Sep 30, 2009 at 16:18 UTC
I want to get $path from $target. How I do it in one regexp? `# How to optimize?` Those are two different questions, and are likely to have different answers. I would leave it as is, but if you insist on using one regexp, I guess (haven't tried it), something like: `$path =~ s/(?:[^/]\?.$)\|(?:[^/]+$)/;` [download]	[reply] [d/l] [select]
Re: How to replace in one regexp by vitoco (Hermit) on Sep 30, 2009 at 16:27 UTC
`my $path = $1 if $target =~ m!^([^?]?/)?([^/\?])(\?(.*))?$!;` [download] `$1` : path `$2` : file or script name `$3` : query string `$4` : query string without leading "?"	[reply] [d/l] [select]
Re: How to replace in one regexp by vitoco (Hermit) on Oct 01, 2009 at 21:46 UTC
`#$path =~ s{\?.$}{}; #$path =~ s{[^/]+$}{}; $path =~ s{[^/\?](\?.*)?$}{};` [download]	[reply] [d/l]