Capturing the path to a file(s) 2

[andyv12]

If I invoke a Perl script that takes a pathname (with wildcard) as its only argument, how do I capture the path without the wildcard?

Ex:

perlscript.pl /Users/andyv/ps/*

How do I capture the full path minus the wildcard into a variable (just /Users/andyv/ps/)??

Thanks!

AndyV

 

[rob51383]

This will remove the *

#!/usr/bin/perl
print "Content-type: text/html\n\n";

$var1 = "perlscript.pl /Users/andyv/ps/*";

$var1=~s/\*//;

print $var1;

 

[arn0ld]

for my information, how do you get the '*' into your script?
All the shells I am familiar with would replace the * with everything under /Users/andyv/ps (unless /Users/andyv/ps did not exist )

 

[mikevh]

Good point, arn0ld.

use File::Basename;

my $dirname = dirname($ARGV[0]);
print "$dirname\n";