How can I capture two columns of ls output in a bash script

Question

I want to create a script to delete any directories with numeric groups.
What I have so far captures and analyzes the group:

#!/bin/bash
re='^[0-9]+$'
for x in `/usr/bin/ls -l /var/indexes | sort -k3,3 | tr -s " " | cut -f3 -d " "`
do
if [[ $x =~ $re ]] ; then
   echo "found one $x"
fi
done

The problem is, the actual name of the directory is in another column.
Here is one line of the ls output:

drwxr-sr-x. 3                     3966 root 4096 Oct 20 16:29 dirname

So I need to iterate through the ls results and capture both the group (in this case 3966) and the directory name (in this case dirname) so I can form an rm -rf command to remove it when I find a blank one.
How would I do that?
Thanks in advance.

Answer 1

Wouldn't it be easier to use find with the -nogroup flag? For example:

find /var/indexes -nogroup

If you want to base the script around ls then awk is tool I'd use to select columns.

Answer 2

You don't really want to do this in bash, but because you asked, this is how you could split it up.

#!/bin/bash
re='^[0-9]+$'
for x in `/usr/bin/ls -l --time-style=+'%s' /var/indexes | sort -k3,3 | awk '{print $3 $7}'`
do
read -A nm <<< "$x"
if [[ ${nm[0]} =~ $re ]] ; then
   echo "found one " ${nm[1]}
fi
done

It's easier in awk like so:

/usr/bin/ls -l --time-style=+'%s' /var/indexes | sort -k3,3 | awk '($3 ~ /^[0-9]+$/){print $7}'

Really, the right solution is find w/ -nogroup as suggested by Paul H..