Re: Quick regex substitution question

Here's a look-ahead assertion that ought to do what you need. It looks for a ';' with anything after it between another ';', and replaces the first one. When it hits the last one, there's no more characters before another ';', so it has nothing else to substitute.

EDIT: It does not hit end-of-line as I previously stated. This example will leave any chars following the last ';', if there's no other ';' before EOL.

#!/usr/bin/perl

use warnings;
use strict;

my $string = "I have; in my string; replace all but last one;";

$string =~ s/;(?=.*;)//g;

print $string;
[download]

Output:

./r.pl
I have in my string replace all but last one;
[download]

-stevieb

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Re^2: Quick regex substitution question by stevieb (Canon) on May 29, 2015 at 21:15 UTC
For the sake of people in the future reading this thread, you can use look ahead/behind to split() something, and retain the thing you split on. The following split() splits on either whitespace or nothing with a '.' before it. The whitespace, if present, is thrown away, but the dot is retained within each split() element. `#!/usr/bin/perl my $string = "I have. In my string.Replace all but last one."; @parts = split(/(?<=\.)\s*/, $string); print "$_\n" for @parts; __END__ ./split.pl I have. In my string. Replace all but last one.` [download] -stevieb	[reply] [d/l]
Re^2: Quick regex substitution question by jmmach80 (Initiate) on May 29, 2015 at 14:21 UTC
Sweet, that looks like it did what I'm looking for. Thanks for the help!	[reply]