Good
point.
maby we have only part of informations and
$ARGV[0] is use for anything else ... but if there is only one argument, it's
$ARGV[0] of course.
----- Original Message -----
Sent: Wednesday, November 19, 2003 3:14
PM
Subject: RE: command line arg
It
should be :
$exy=$ARGV[0];
I am trying to pass a file extension as a command line argument, then
to search all directories for that partcular extension, but have problem
getting it to work. Could you please advise.
$ext = $ARGV[1];
...
...
...
&bad_files($entry,$size) if($entry =~ /\$ext$/);
Thanks
Terry
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