Sounds like you tried to get the filename into the $file variable by parsing it out of a directory listing. For example, you might have done something like
If you are going to do that, you'll need to improve your parsing. ;-)for my $file (`ls -l /some/dir`) { ... }
If you show us the code next time, you'll probably get a more thorough response.
-sauoq "My two cents aren't worth a dime.";
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