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pinga123

05-20-2011 02:27 AM

Scripting help needed

For getting individual entries from $PATH variable i m using following scrip let.

Code:

#!/bin/bash



COUNT=`echo "$PATH" | awk -F ":" '{print NF}'`

while [ "$COUNT" -ge 1 ]

do

echo "$PATH" | awk -v m=$COUNT -F ":" '{print "ls -ld " $m}'|sh

COUNT=`expr $COUNT - 1`

done

Is there any other way of doing it ?

colucix

05-20-2011 02:35 AM

Yes. You can either do anything from awk or use pure bash. The latter will look like this:

Code:

#!/bin/bash

OLD_IFS="$IFS"



IFS=":"

for dir in $PATH

do

  ls -ld $dir

done



IFS="$OLD_IFS"

This take advantage from the fact you can change the input field separator, so that the for loop iterates over the colon separated list of directories. The OLD_IFS stuff serves to save and restore the default value before and after the loop respectively.

catkin

05-20-2011 03:04 AM

Hello colucix :)

As a variation on the same (at the expense of using a subshell, avoids need to save and restore IFS):

Code:

( export IFS=':'; for dir in $PATH; do ls -ld $dir; done )

EDIT:

It could more simply be:

Code:

( IFS=':'; for dir in $PATH; do ls -ld $dir; done )

colucix

05-20-2011 03:15 AM

Hello catkin! :)

That's a nice idea (I always hated this OLD_IFS stuff)! Something to take in mind as future reference.

grail

05-20-2011 04:21 AM

How about we just make it easy and leave IFS alone :)

Code:

for dir in ${PATH//:/ }; do ls -ld $dir; done

colucix

05-20-2011 04:27 AM

Hi grail! :) That's right! Another nice solution.

grail

05-20-2011 04:50 AM

Actually, I assume something else must be going to be done in the loop as the substituion could just be used with the ls on its own:

Code:

ls -ld ${PATH//:/ }

catkin

05-20-2011 06:15 AM

Quote:

Originally Posted by grail (Post 4361701)

Actually, I assume something else must be going to be done in the loop as the substituion could just be used with the ls on its own:

Code:

ls -ld ${PATH//:/ }

Neat :)

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