substituting a string in place of variable in shell

Question

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

score 4 · Accepted Answer · answered Aug 13 '11 at 13:11

4

The syntax for this is:

${!VAR}

Example:

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

answered Aug 13 '11 at 13:11

Mat

Nice! Is there a way to use this for assignment, i.e. some replacement for `eval $1=foo` ? – user123444555621 Aug 13 '11 at 16:32
1

See this article for variations on that: [Indirect assignment](http://bashify.com/?Useful_Techniques:Indirect_Variables:Indirect_Assignment) – Mat Aug 13 '11 at 16:57
@Mat Good stuff. But this is even better: http://stackoverflow.com/a/13717788/27862 ;) – user123444555621 Dec 20 '12 at 07:48

score 1 · Answer 2 · answered Aug 13 '11 at 13:16

1

All you should do:

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

answered Aug 13 '11 at 13:16

pkk

2 Answers2