Put grep output inside variable from a loop

Question

I have CentOS and this bash script:

#!/bin/sh
files=$( ls /vps_backups/site )
counter=0
for i in $files ; do
echo $i | grep -o -P '(?<=-).*(?=.tar)'
let counter=$counter+1
done

In the site folder I have compressed backups with the following names :

    site-081916.tar.gz
    site-082016.tar.gz
    site-082116.tar.gz
    ...

The code above prints : 081916 082016 082116

I want to put each extracted date to a variable so I replaced this line

echo $i | grep -o -P '(?<=-).*(?=.tar)'

with this :

dt=$($i | grep -o -P '(?<=-).*(?=.tar)')
echo $dt

however I get this error :

./test.sh: line 6: site-090316.tar.gz: command not found

Any help please?

Thanks

J Earls · Accepted Answer · 2016-09-05 21:44:42Z

2

you still need the echo inside the $(...):

dt=$(echo $i | grep -o -P '(?<=-).*(?=.tar)')

answered Sep 5, 2016 at 21:44

J Earls

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2 Comments

oussama kamal Over a year ago

Well, that's a fast and perfect answer for my issue, I'm new to bash, thank you.

J Earls Over a year ago

for what it's worth: you can make this more efficient by moving the grep into the $files definition: files=$( ls /vps_backups/site | grep -o -P '(?<=-).*(?=.tar)' ) - that way, files will be just the filename parts you're looking for already.

chepner · Accepted Answer · 2016-09-05 23:03:37Z

1

Don't use ls in a script. Use a shell pattern instead. Also, you don't need to use grep; bash has a built-in regular expression operator.

#!/bin/bash
files=$( /vps_backups/site/* )
counter=0
for i in "${files[@]#/vps_backups/site/}" ; do
  [[ $i =~ -(.*).tar.gz ]] && dt=${BASH_REMATCH[1]}
  counter=$((counter + 1))
done

answered Sep 5, 2016 at 23:03

chepner

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