Regex to replace all ocurrences of a given character, ONLY after a given match.

Question

For the sake of simplicity, let's say that we have input strings with this format:

*text1*|*text2*

so, I want to leave text1 alone, and remove all spaces in text2

This could be easy if we didn't have text1, a simple search and replace like this one would do:

%s/\s//g

but in this context I don't know what to do.

I tried with something like:

%s/\(.*|\S*\).\(.*\)/\1\2/g

which works, but removing only the first character, I mean, this should be run on the same line one time for each offending space.

So, a preferred restriction, is to solve this with only one search and replace. And, although I used vim syntax, use the regex flavor you're most comfortable with to answer, I mean, maybe you need some functionality only offered by Perl.

Edit: My solution for vim:

%s:\(|.*\)\@<=\s::g

Answer 1

One way, in perl:

s/(^.*\||(?=\s))\s*/$1/g

Certainly much greater efficiency is possible if you allow more than just one search and replace.

Answer 2

So you have a string with one pipe (|) in it, and you want to replace only those spaces that don't precede the pipe?

s/\s+(?![^|]*\|)//g

Answer 3

you might try embedding perl code in regex (using (?{...}) syntax), which is however rather an experimental feature and might not work or even be available in your scenario.

this

s/(.*?\|)(.*)(?{ $x = $2; $x =~ s:\s::g })/$1$x/

should theoretically work, but i got "Out of memory!" failure, which can be fixed replacing '\s' with a space:

s/(.*?\|)(.*)(?{ $x = $2; $x =~ s: ::g })/$1$x/