How can I build regexp with "not" assertion?

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

Dear Monks, I have to build a substitution regexp, but this case is too hard for me.
We have a data:

 
1stline....pattern1.......pattern1.....
2ndline....pattern1.....pattern2...pattern1...
1a line....pattern1.......pattern1.....
2a line....pattern1........pattern2...pattern1.....
[download]

"\n" and "." doesn't matter (so I can't split it into many lines and process separately). line numbers just for reference, and real data may mix 1st and 2nd type of lines in random way.

what we must get:

 
1stline....pattern1.......pattern1.....
2ndline....subst...
1a line....pattern1.......pattern1.....
2a line....subst.....
[download]

I realize I should do it with
s/pattern1(not-pattern1)*?pattern2(not-pattern1)*?pattern1/subst/isg
but hung.
What I already tried:

pattern1([^p][^a][^t]...[^1])*?pattern2([^p][^a][^t]...[^1])*?pattern1
pattern1.(pattern1){0}*?pattern2.(pattern1){0}*?pattern1
[download]

and so on.
Of couse tried:
O'Reilly Perl bookshelf
man perlretut
Also I tried to search, but (probably because of my English) I didn't find anything helpful.

perl, v5.8.6, Linux.

Please, help. Thank you very much.

Comment on How can I build regexp with "not" assertion? Select or Download Code

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Re: How can I build regexp with "not" assertion? by Erez (Priest) on Feb 17, 2008 at 14:20 UTC
Perl supports negative lookbehing and negative lookahead, filed under EXTENDED CONSTRUCTS of perlreref and Extended Patterns of perlre. Lookbehing is slightly limited in scope (i.e no quantifiers), but lookahead isn't. Also, in regex, you might be better off stating what you are looking for and unless the rest UPDATE:Fixed perldoc links, thanks kyle! Software speaks in tongues of man. Stop saying 'script'. Stop saying 'line-noise'. We have nothing to lose but our metaphores.	[reply]
Re^2: How can I build regexp with "not" assertion? by putnik (Novice) on Feb 17, 2008 at 19:10 UTC
Thank you, I'll go to read, and be back if I miss something :)	[reply]
Re: How can I build regexp with "not" assertion? by parv (Parson) on Feb 17, 2008 at 12:58 UTC
Given your example, if you are trying to do substitutions on all the lines which match all the 2-numbered lines, just ignore the "not parts" ... `$line =~ s/(pattern1).+?pattern2.+?\1/substitute/;` [download] ... Above will work since "pattern2" does not appear (per your example) in any lines identified by the 1s, so they remain unchanged.	[reply] [d/l]
Re^2: How can I build regexp with "not" assertion? by putnik (Novice) on Feb 17, 2008 at 17:36 UTC
Thank you, but sorry, it work in wrong way. I tried this, but got `1stline....subst... 1a line....subst.....` [download] I mean it substituted all in parenthesis this way: `1stline....(pattern1.......pattern1..... 2ndline....pattern1.....pattern2...pattern1)...` [download] I CAN'T split it in lines, i must treat all dada as a single line because \n may appear in any place	[reply] [d/l] [select]
Re: How can I build regexp with "not" assertion? by johngg (Canon) on Feb 17, 2008 at 17:47 UTC
Rather than using a pure regex solution, you can find all of the starting offsets of `pattern1` and `pattern2` and then use List::Util`::first` (along with reverse) to find which pairs of `pattern1` bracket each `pattern2`. Once you have that you can use substr to replace the text working from the end of the string backwards so the offsets aren't invalidated as the string changes. use strict; use warnings; use List::Util q{first}; my $patt1 = q{ABC}; my $patt2 = q{XYZ}; my $replace = q{999999}; my $string = <<EOD; kjdfjdfXYZewfkfABClkjfef sahasjABCsjhksfhABCsjsjfs oreweouABCkerjeXYZewfkfABClkjfef xcvmvbbbvABCdjfABCjsdjfsdf jjnnjfABDjfXYZdjdjABCjfdkjfABClsfj isosiXYZcsfsjfABChfdhgfABCyeryerXYZffjfs EOD print $string, q{-} x 25, qq{\n}; my @patt1Posns = (); push @patt1Posns, ( pos( $string ) - length $patt1 ) while $string =~ m{\Q$patt1}g; my @patt2Posns = (); push @patt2Posns, ( pos( $string ) - length $patt2 ) while $string =~ m{\Q$patt2}g; my @substituteSets = (); foreach my $patt2Posn ( @patt2Posns ) { my $start = first { $_ < $patt2Posn } reverse @patt1Posns or next; my $end = first { $_ > $patt2Posn } @patt1Posns or next; push @substituteSets, [ $start, $end - $start + length $patt1 ]; } for my $raSubstituteSet ( reverse @substituteSets ) { substr $string, $raSubstituteSet->[ 0 ], $raSubstituteSet->[ 1 ], $replace; } print $string, q{-} x 25, qq{\n}; [download] This produces `kjdfjdfXYZewfkfABClkjfef sahasjABCsjhksfhABCsjsjfs oreweouABCkerjeXYZewfkfABClkjfef xcvmvbbbvABCdjfABCjsdjfsdf jjnnjfABDjfXYZdjdjABCjfdkjfABClsfj isosiXYZcsfsjfABChfdhgfABCyeryerXYZffjfs ------------------------- kjdfjdfXYZewfkfABClkjfef sahasjABCsjhksfhABCsjsjfs oreweou999999lkjfef xcvmvbbbvABCdjf999999jfdkjf999999hfdhgfABCyeryerXYZffjfs -------------------------` [download] I hope this is of use. Cheers, JohnGG Update: Fixed typo. s/and/as/	[reply] [d/l] [select]
Re: How can I build regexp with "not" assertion? by ikegami (Patriarch) on Feb 17, 2008 at 18:29 UTC
This will probably work, depending on what `pattern1` and `pattern2` really are: `/ pattern1 (?:(?!pattern1).)* pattern2 (?:(?!pattern1).)* pattern1 /x` [download]	[reply] [d/l] [select]
Re^2: How can I build regexp with "not" assertion? by putnik (Novice) on Feb 22, 2008 at 14:09 UTC
THANK YOU! THANK YOU! THANK YOU! It greatly work. Googd Luck!	[reply]
Re: How can I build regexp with "not" assertion? by nikhil.patil (Sexton) on Feb 18, 2008 at 07:00 UTC
I think what you need is: `$line =~ s/(pattern1)((?!pattern1).)+pattern2.+?\1/substitute/gs;` [download] It worked well on Perl v5.8.8	[reply] [d/l]
Re^2: How can I build regexp with "not" assertion? by putnik (Novice) on Feb 22, 2008 at 14:15 UTC
Thank you, your solution work too :) I didn't check it tightly, but it's look suitable.	[reply]