CODEWITHSUNDEEP
regex
Qqwy
Qqwy

Reputation: 5629

Regular Expressions: Remove character before match

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.

Upvotes: 0

Views: 945

Answers (2)

Birei
Birei

Reputation: 36262

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

Upvotes: 1

coconut
coconut

Reputation: 1101

As far as I know, regex match things. They don't delete or modify input.

A pretty straightforward algorithm would be:

1 - Read your input counting the characters until you get to a [removebefore].

2 - When you get to [removebefore], count the number of [removebefore] in another variable.

3- Subtract the number of [removebefore] from the number of characters and save it in a variable, let's call it "n".

4 - Go again through your input, printing the first "n" number of characters.

There is some error handling to be done in this algorithm, but that's basically it.

Upvotes: 0

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