Question

I was wondering if there was a way to use Regular expressions to remove n occurrences of characters right before n matches of [removebefore]

Might be unclear, but an example explains everything:

This is a teststring[removebefore][removebefore][removebefore]. blablabla[removebefore][removebefore]

should be changed into

This is a teststr. blablab

Of course this can be done by matching and replacing /.\[removebefore\]/ a dozen of times, but is there a way to do this in one regexp?

EDIT:
I was trying to do this in PHP. And indeed I am searching for a regular expression that matches [anychar]{n}\[removebefore\]{n} where n is the number of removebefore’s.

Answer 1

One possible solution using perl

Regular expression:

s<([^\[]+)((?:\[removebefore\])+)><substr( $1, 0, length($1) - scalar(split(/]\[/, $2)) )>ge

Test:

Content of script.pl:

use warnings;
use strict;

while ( <DATA> ) {
    s<([^\[]+)((?:\[removebefore\])+)><substr( $1, 0, length($1) - scalar(split(/]\[/, $2)) )>ge;
    print;
}   

__DATA__
This is a teststring[removebefore][removebefore][removebefore]. blablabla[removebefore][removebefore]

Running script:

perl script.pl

And result:

This is a teststr. blablab