So... what? you want to get a scalar var containing the filename? So you can print it out or something?

If that's what you want, There's A Lot More Than One Way To Do It. Don't let anyone around here hear you say "Perl doesn't let me search and replace..." ! If anyone can search and replace it's Perl.

Assuming you have 
my $filepath = 'c:\somedir\file.ext'

[download]
and you want to get 
$filename eq 'file.ext'

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then 
$filepath =~ /([\w|\.]*)$/;
my $filename = $1;

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Oughta do it - i.e. capture the longest possible string at the end of the filepath that is either a word character (alpha-num or _) or a dot. Or if you think there might be non-word characters in the file name, $filepath =~ /([^\\|^\/]*)$/; - capture the longest string at the end that doesn't have either a "\" or a "/".

I'm no regex guru, so there may be obscure pitfalls in either of these, or a neater way to do it... but that's what I'd do.

Of course, I may have completely mis-understood your question again :)

§ George Sherston

In reply to Re: Re: Re: Re: Re: convert windows path by George_Sherston
in thread convert windows path by marauder

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