Parsing and Printing $PATH Using Unix

Question

I've placed my PATH in a text file and would like to print each path on a newline using a simple command in UNIX.

I've found a long way to do it that goes like this...

cat Path.txt | awk -F\; '{print $1"\n", $2"\n", ... }'

This however seems inefficient so I know there must be a way to quickly print out my results on new lines each time without having to manually call each field separated by the delimiter.

The path separator on Unix is colon `:` and not semi-colon `;` as set in the `awk` command in the question. — Jonathan Leffler, Oct 24 '12 at 00:07

score 17 · Accepted Answer · answered Oct 23 '12 at 23:45

17

Yet another way:

echo $PATH | tr : '\n'

or:

tr : '\n' <Path.txt

answered Oct 23 '12 at 23:45

Olaf Dietsche

72,253
8
102
198

Since I've been ragging on other people for bashisms, I feel like I have to point out that some systems ship a `tr` that doesn't understand `'\n'`. '' works, though. – zwol Oct 23 '12 at 23:55
In case anyone else wonders: PATH entries are not allowed to contain colons. Ie. there's no escape mechanism: http://stackoverflow.com/a/29213487/1517969 – olejorgenb Aug 06 '16 at 10:59

score 2 · Answer 2 · answered Oct 24 '12 at 03:26

2

The tr solution is the right one but if you were going to use awk then there'd be no need for a loop:

$ echo "$PATH"
/usr/local/bin:/usr/bin:/cygdrive/c/winnt/system32:/cygdrive/c/winnt

$ echo "$PATH" | awk -F: -v OFS="\n" '$1=$1'
/usr/local/bin
/usr/bin
/cygdrive/c/winnt/system32
/cygdrive/c/winnt

answered Oct 24 '12 at 03:26

Ed Morton

188,023
17
78
185

Just to mention gawk and mawk have native windows binary, work well with cmd.exe, cheers! – MeaCulpa Oct 24 '12 at 05:02
what does the OFS mean in the above comment? – Nick Oct 29 '12 at 22:59
It's the Output Field Separator. I'm telling awk to take a list fields separated by colons (FS = input Field Separator) and print them separated by newlines (OFS = Output Field Separator) instead. – Ed Morton Oct 29 '12 at 23:18

Gilles Quénot · Answer 3 · 2012-10-24T01:57:09.183

0

With Perl for UNIX/UNIX-likes :

echo $PATH | perl -F: -ane '{print join "\n", @F}'

With any OSes (tested on Windows XP, Linux, Minix, Solaris):

my $sep;
my $path;

if ($^O =~ /^MS/) {
    $sep = ";";
    $path = "Path";
}
else {
     $sep = ":";
     $path = "PATH";
}

print join "\n", split $sep, $ENV{$path} . "\n";

If using bash for Unix, try the following code :

printf '%s\n' ${PATH//:/ }

This use bash parameter expansion

edited Oct 24 '12 at 01:57

answered Oct 23 '12 at 23:43

Gilles Quénot

173,512
41
224
223

The only virtue of shell is its extreme portability. If you can't do something in shell using only the facilities that existed in Unix98, you are better off switching to a real programming language than making use of unportable shell extensions. – zwol Oct 23 '12 at 23:47
@Zack, if you need portability, stop using shell, and take a try to python, ruby or perl. – Gilles Quénot Oct 23 '12 at 23:58
But that's exactly my point! The moment you need an unportable shell feature is the moment to stop using shell! – zwol Oct 24 '12 at 00:01
Added another one Perl solution. – Gilles Quénot Oct 24 '12 at 00:04

score 0 · Answer 4 · answered Oct 24 '12 at 00:13

I have a Perl script that I use for this:

#!/usr/bin/env perl
#
#   "@(#)$Id: echopath.pl,v 1.8 2011/08/22 22:15:53 jleffler Exp $"
#
#   Print the components of a PATH variable one per line.
#   If there are no colons in the arguments, assume that they are
#   the names of environment variables.

use strict;
use warnings;

@ARGV = $ENV{PATH} unless @ARGV;

foreach my $arg (@ARGV)
{
    my $var = $arg;
    $var = $ENV{$arg} if $arg =~ /^[A-Za-z_][A-Za-z_0-9]*$/;
    $var = $arg unless $var;
    my @lst = split /:/, $var;
    foreach my $val (@lst)
    {
        print "$val\n";
    }
}

I invoke it like:

echopath $PATH
echopath PATH
echopath LD_LIBRARY_PATH
echopath CDPATH
echopath MANPATH
echopath $CLASSPATH

etc. You can specify the variable name, or the value of the variable; it works both ways.

score 0 · Answer 5 · answered Oct 24 '12 at 04:54

0

awk:

echo $PATH|awk -F: '{gsub(/:/,"\n");print}'

perl:

echo $PATH|perl -F: -lane 'foreach(@F){print $_}'

answered Oct 24 '12 at 04:54

Vijay

65,327
90
227
319

score 0 · Answer 6 · answered Oct 24 '12 at 05:00

0

for AWK, in addition to:

echo $PATH | awk -vFS=':' -vOFS='\n' '$1=$1'

You can:

echo $PATH | awk -vRS=':' '1'

answered Oct 24 '12 at 05:00

MeaCulpa

881
1
6
14

Parsing and Printing $PATH Using Unix

6 Answers6