How can I increment a var in a substitution?

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

Hi, I just read [id://20361|How would I replace the string 'Red1' with $Red1?]
But I still can't figure this out. Here's the code:

$n = 0;
$html_file_slurped_to_a_scalar =~ s/<img .*?>/ŤTOKEN_$n++ť/gis;
[download]

This turns all img links into ŤTOKEN_0++ť; $n interpolotes but doesn't increment.
If I add the /e modifier, the complier chokes when it tries to evaulate Ť

Unrecognized character \xAB at /path/foo line 25.
[download]

Maybe I'm trying to do too much but I'd like to avoid the following loop if I can:

evaluate $html_file_slurped_to_a_scalar
change the first image tag to ŤTOKEN_$nť
increment $n

Any suggestions?

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Re: How can I increment a var in a substitution? by suaveant (Parson) on Aug 09, 2001 at 23:11 UTC
the e modifier means evaluate right side of substitution as perl code... so you need `/'<<TOKEN_'.$n++.'>>'/` to make it into perl code that makes the string you want to substitute... you could also do... `$html_file_slurped_to_a_scalar =~ s/<img .*?>/ŤTOKEN_${\($n++)}ť/gis;` [download] which is a little trick to put code into an interpolated string - Ant - Some of my best work - Fish Dinner	[reply] [d/l] [select]