Bash scripting problem

mastertaf · 09-19-2008, 08:37 AM

Hello,

this is my first post so i hope that am doint this right. I have a fairly simple problem, or it should be. I searched on google wasnt able to find an answer that i could understand. I am making a script to execute some common task on my linux servers and within the script i would like to identify the version of the os. I only use CentOS, and on red hat systems there is a file name /etc/redhat-release that contains the info i need. here is part of the code that is not working:

Code:

VERSION= cat /etc/redhat-release | awk '{print $3}'

if [ $VERSION > 5 ];then
        echo "Centos version 4"
else
        echo "Centos Version 5"
fi

I have, i think, identified part of the problem. The variable value is a string so i guess the logical test cannot work...and since i am not sure how to convert the value to an integer..

Thank you in advance.

pixellany · 09-19-2008, 08:57 AM

Maybe try it using something like:

if [ $VERSION -gt 5 ]
or
if [[ $VERSION > 5 ]]

Some of the syntax of BASH is pretty inscrutable---the best tool I have found is to keep "ABS" on my computer and use it for searches. (Print it only if you really hate trees....)
http://tldp.org/LDP/abs/html/

pixellany · 09-19-2008, 08:58 AM

PS:
Everything is stored as strings--BASH treats strings as numbers depending on context

mastertaf · 09-19-2008, 09:03 AM

Thank you pixellany your solution worked perfectly. it was simply a question of adding the extra [].

I will keep ABS on my copmuter from now on!

Code:

#!/bin/bash
VERSION=`cat /etc/redhat-release | awk '{print $3}'`
echo $VERSION
if [[ $VERSION < 5 ]];then
        echo "Centos version 4"
else
        echo "Centos Version 5"
fi

234107 · 09-19-2008, 09:04 AM

Quote:

Originally Posted by mastertaf

Hello,

this is my first post so i hope that am doint this right. I have a fairly simple problem, or it should be. I searched on google wasnt able to find an answer that i could understand. I am making a script to execute some common task on my linux servers and within the script i would like to identify the version of the os. I only use CentOS, and on red hat systems there is a file name /etc/redhat-release that contains the info i need. here is part of the code that is not working:

Code:

VERSION= cat /etc/redhat-release | awk '{print $3}'

if [ $VERSION > 5 ];then
        echo "Centos version 4"
else
        echo "Centos Version 5"
fi

I have, i think, identified part of the problem. The variable value is a string so i guess the logical test cannot work...and since i am not sure how to convert the value to an integer..

Thank you in advance.

Try this...

Code:

VERSION=$(cat /etc/redhat-release | awk '{print $3}')

if [ $VERSION > 5 ];then
        echo "Centos version 4"
else
        echo "Centos Version 5"
fi