Using a BASH integer variable in an awk command

Question

I'm trying to run a command on every entry in a line of text:

#! /bin/bash

range=$(ls -s /home/user/Pictures/Wallpapers | wc -l)

for (( i=0; i<$range; i++ ))
do  
   d=$(ls /home/user/Pictures/Wallpapers | paste -sd " " | awk -F' ' '{ print $i }')
echo $d
done

But when I run it, it simply prints out the entire line of file entries with every iteration of the loop. I think it's because the "i" is not being interpreted since it's technically in between single quotes, and I can't figure out how to prevent it from doing this.

Thanks.

What are you trying to do? Print out the numbers from 0 to $range, $range times? — Joni
– Joni, Commented Oct 2, 2013 at 22:17
I'm trying to print out every filename in a directory, range is the number of files in the directory, so I want to print out a file name $range times. — KingOfEphrya
– KingOfEphrya, Commented Oct 2, 2013 at 22:23

EverythingRightPlace · Accepted Answer · 2013-10-02 22:51:26Z

3

You can pass variables to awk with -v argument like in:

d=$(ls /home/user/Pictures/Wallpapers | paste -sd " " | awk -v var="$i" -F' ' '{ print var }')

answered Oct 2, 2013 at 22:51

EverythingRightPlace

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Comments

Joni · Accepted Answer · 2013-10-02 22:27:43Z

1

To print the names of files in a directory repeated as many times as there are files in the directory, you can use:

for x /home/user/Pictures/Wallpapers/*; do
   ls /home/user/Pictures/Wallpapers/
done

or

for x /home/user/Pictures/Wallpapers/*; do
  for y /home/user/Pictures/Wallpapers/*; do
    echo $x
  done
done

answered Oct 2, 2013 at 22:27

Joni

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Comments

Cole Tierney · Accepted Answer · 2013-10-03 00:26:17Z

0

for f in /home/user/Pictures/Wallpapers/*; do echo $f; done

answered Oct 3, 2013 at 0:26

Cole Tierney

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