BASH get variable according to name of another variable

Shell Programming and Scripting
1

Just started BASH scripting, and I tried to make a script 'args' to display all of the arguments that I give to it.

#!/bin/bash
if [ $# -eq 0 ]
then
    echo "No arguments specified."
fi
val=
for ((i=1; i <= $# ; i++))
do
    eval "\$val=\$$i"
    echo "Argument number $i is $var."
done

However when I run 'args abc defgh' I get the following output:

What's wrong with the eval statement?

2

Hi,

this cannot work:

eval "\$val=\$$i"

Under bash it is the easiest to use:

    var=${!i}

Or on other shells:

    var=`eval echo \$$i`

HTH Chris

3

Here is one way of doing what you want to do.

for ((i=1; i <= $#; i++))
do
    eval val='${'$i'}'
    echo "Argument number $i: $val"
done

$ ./testcode one two three
Argument number 1: one
Argument number 2: two
Argument number 3: three
$
4

Thanks for your suggestions! I've got it working now! :b: