How to get the Directory Path

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

I am trying to chop the directory path... What my code does is looks in a file to get the file names and ftp the files to the location in the path. My path varies so I want to get it into a variable and cd to that dir and move the file there. In the path c:/Programfile/Perl/test/temp.txt How can I get c:/Programfile/Perl/test into a variable? Any ideas.... Thanks

Comment on How to get the Directory Path

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Re: How to get the Directory Path by ikegami (Patriarch) on Sep 22, 2005 at 19:36 UTC
The portable way: `use File::Spec (); my $path = 'c:/Programfile/Perl/test/temp.txt'; my ($volume, $directories, $file) = File::Spec->splitpath($path); my $dir = File::Spec->catpath($volume, $directories, ''); # Optional: $dir = File::Spec->canonpath($dir); print("$dir\n");` [download] Update: Fixed typo in var name.	[reply] [d/l]
Re^2: How to get the Directory Path by perl_99_monk (Novice) on Sep 23, 2005 at 16:39 UTC
Thanks a lot... It worked..	[reply]
Re: How to get the Directory Path by davidrw (Prior) on Sep 22, 2005 at 20:25 UTC
Check out File::Basename : `use File::Basename; my $filename = 'c:/Programfile/Perl/test/temp.txt'; my $dirname = dirname($filename);` [download]	[reply] [d/l]