Using indirect variable references in a Bash script

Question

I have lots of old log files in a directory, with apps logs generated every month to this directory. I want to use a script to remove only the last month old logs.

This is my file logsDir , these files are 0kb, i have created for question understanding

[root@dasari9 logDir]# ls -ltr
total 136
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram9.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram8.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram7.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram6.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram69.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram68.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram67.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram66.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram5.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram59.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram58.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram57.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram56.logs
 -rw-r--r-- 1 root root 0 Mar 20 12:22 ram4.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram49.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram48.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram47.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram46.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram3.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram39.logs
 -rw-r--r-- 1 root root 0 Mar 20 12:22 ram38.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram37.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram36.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram2.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram29.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram28.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram27.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram26.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram1.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram19.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram18.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram17.logs
-rw-r--r-- 1 root root 0 Mar 20 12:22 ram16.logs

This is my logs.sh:

#!/bin/bash

Jan=Dec Feb=Jan Mar=Feb Apr=Mar May=Apr Jun=May Jul=Jun Aug=Jul Sep=Aug Oct=Sep Nov=Oct Dev=Nov

DIR=/logDir
cd $DIR
if [ $DIR -eq $PWD ]; then 
PWD=$(pwd)
cmd=$(date |awk '{print $2}') 
dat=$(ls -ltr |grep $cmd |awk '{print $9}')
rm -rf $dat
else
echo " enterend in the wrong dir"
fi

but it's failing to get that variable value Apr=Mar

This is the output, I'm trying to delete Apr files

[root@dasari9 /]# bash -x log.sh 
+ Jan=Dec
+ Feb=Jan
+ Mar=Feb
+ Apr=Mar
+ May=Apr
+ Jun=May
+ Jul=Jun
+ Aug=Jul
+ Sep=Aug
+ Oct=Sep
+ Nov=Oct
+ Dev=Nov
+ DIR=/logDir
+ cd /logDir
++ pwd
+ PWD=/logDir
+ '[' /logDir == /logDir ']'
++ date
++ awk '{print $2}'
+ cmd=Apr
++ ls -ltr
++ grep '$Apr'
++ awk '{print $9}'
+ dat=
+ rm -rf

Without knowing the content of the directory, it's hard to understand what's happening here. What happens when you run ls -ltr | grep Mar in the directory? Does the output show some of the files, and does the output have 9 columns? — nwinkler
– nwinkler, Commented Jun 16, 2015 at 11:27
Better post the code in shellcheck.net for an initial safety check. Also, Jan=Dec Feb=Jan Mar=Feb Apr=Mar May=Apr Jun=May Jul=Jun Aug=Jul Sep=Aug Oct=Sep Nov=Oct Dev=Nov looks quite weird. — fedorqui
– fedorqui, Commented Jun 16, 2015 at 11:32
running with ls -ltr | grep Mar |awk '{print $9}' , it gives all march file names only , with out all fields like type of file , time,date..etc..I'm adding files also now — radas
– radas, Commented Jun 16, 2015 at 11:33
Why don't you use find for this? Something like find $DIR -mtime +30 will find you files that have not been modified for more than 30 days. — Filipe Gonçalves
– Filipe Gonçalves, Commented Jun 16, 2015 at 11:34
Or better yet; this mostly reduces your script to one line: find . -mtime +30 -exec echo 'rm {}' \; | bash -x — Filipe Gonçalves
– Filipe Gonçalves, Commented Jun 16, 2015 at 11:41

nwinkler · Accepted Answer · 2015-06-16 11:45:17Z

1

It seems like you're looking for an indirect variable reference. Your $cmd variable points to the variable $Apr and you want to use the value of $Apr.

Here's how you can do this (just the relevant lines):

cmd=$(date | awk '{print $2}')
eval month=\$$cmd
dat=$(ls -ltr | grep $month | awk '{print $9}')

The eval command resolves the indirect reference created by \$$cmd.

More details here: http://www.tldp.org/LDP/abs/html/ivr.html

answered Jun 16, 2015 at 11:45

nwinkler

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