CODEWITHSUNDEEP
linuxbashshellvariablesscripting
raj
raj

Reputation: 3811

substituting a string in place of variable in shell

I pass a string as an argument to a shell script. and the shell script should tell me if the passed argument is a variable

something like this

if [ ! -z ${$1} ] ; then  
echo yes! $1 is a variable and its value is ${$1}  
fi

but this gives me bad substitution err..

I definitely know i'm missing something.. help me out!

Eg usage: 

$ myscript.sh HOME  
yes! HOME is a variable and its value is /home/raj

Upvotes: 4

Views: 238

Answers (2)

pkk
pkk

Reputation: 3691

Found it here: http://www.linuxquestions.org/questions/programming-9/bash-how-to-get-variable-name-from-variable-274718/

All you should do:

if [ ! -z ${!1} ]; then
    echo yes $1 is a variable and its value is ${!1}
fi

Upvotes: 1

Mat
Mat

Reputation: 206689

The syntax for this is:

${!VAR}

Example:

$ function hello() { echo ${!1}; }
$ hello HOME
/home/me

Upvotes: 4

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