CODEWITHSUNDEEP
linuxperl
w13rfed
w13rfed

Reputation: 375

Listing Directories on terminal with Perl

I'm trying to write a perl one liner for listing the content of a directory like with the "ls -l" Unix command however I only want to output directories and not files

I have managed to do it using a perl script but I want to try and see if it can be reduced to a single line using a pipe like 

ls -l | grep "something"

or

ls -l | perl -...

the way I have done it so far is this

#!/usr/bin/perl

open(LS_D, "ls -l |");

while(<LS_D>) {
  print if /^d.*/; #print line if line starts with d
}

Also could you tell me why this works with the pipe in "ls -l |" but not with "ls -l"?

Thanks

Upvotes: 0

Views: 499

Answers (1)

zdim
zdim

Reputation: 66964

The long listing of directories can be directly obtained with

ls -ld */

without a need to filter the output.

As for the |, that is how you tell open to open a process and replace its STDIN or STDOUT

For three or more arguments if MODE is |-, the filename is interpreted as a command to which output is to be piped, and if MODE is -|, the filename is interpreted as a command that pipes output to us. In the two-argument (and one-argument) form, one should replace dash (-) with the command. See Using open() for IPC in perlipc for more examples of this.

Without |, so with open(LS_D, "ls -l"), you are trying to open a file, with the name 'ls -l'

The three-argument open, which is recommended with its lexical filehandle, would be

open my $read_fh, '-|', 'ls', '-l';

where the command with arguments is supplied as a list.

Finally, what you have in your Perl script can be done with a one-liner for example as

perl -we'print grep { /^d/ } `ls -l`'

but, as shown above, you can get that directly with ls on the command line.

I should add

  • Always have use warnings; and use strict; at the top of your programs

  • Always check calls like open and you'll know about, and see the reason for, the failure

    open my $fh ...  or die "Can't open ... : $!";

    See Error Variables in perlvar for $!

Upvotes: 4

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