CODEWITHSUNDEEP
bashshellvariablesscripting
IlikePepsi
IlikePepsi

Reputation: 675

Bash scripting - nested variables

I currently encountered a tricky problem on which I couldn't find any solution yet.

I wrote a script like this:

#!/bin/sh 
x=1
while [ "$x" -le $# ]; do
echo "$x"'. Argument is: ' "\$$x"
x="$(( $x + 1 ))"
done

I suggested that the shell would evaluate the expression "\$$x" after expanding the variables to get access to the argument on position x but the output is:

1. Argument is: $1

Please help. Thx in advance.

Upvotes: 0

Views: 2933

Answers (2)

BMW
BMW

Reputation: 45293

Here is the fix

$ cat a.sh
#!/bin/sh

x=1
while [ "$x" -le $# ]; do
echo "$x"'. Argument is: ' "${!x}"    # If you need indirect expansion, use ${!var} is easier way.
x="$(( $x + 1 ))"
done

Test result

$ sh a.sh a b c
1. Argument is:  a
2. Argument is:  b
3. Argument is:  c

Upvotes: 2

drolando
drolando

Reputation: 507

This code should work:

#!/bin/sh
x=0
args=($@)
while [ "$x" -lt $# ]; do
    echo "$x"'. Argument is: ' "${args[${x}]}"
    x="$(( $x + 1 ))"
done

Upvotes: 0

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