How to insert a variable into another variable with bash

Question

I have a problem with a bash script.

I would like to insert a variable into another variable in a bash script, but the risult is not what I should expect.

Here the code

input1="inputnumber1"
input2="inputnumber2"
input3="inputnumber3"
dummy="input"
for i in $(seq 1 3)
do
    toprint=$dummy$i
    echo "$toprint"
done

I would expect this code print the content of the variable $input1, $input2 and $input3, but it just print input1, input2 and input3.

Any suggestion?

Thanks in advance.

user000001 · Accepted Answer · 2013-09-17 07:48:04Z

3

Use an indirect variable reference:

varname="$dummy$i"
toprint="${!varname}"

answered Sep 17, 2013 at 7:48

user000001

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Comments

devnull · Accepted Answer · 2013-09-17 07:48:55Z

2

Say:

echo "${!toprint}";

Also read this. You can also read about indirect expansion in the manual.

answered Sep 17, 2013 at 7:48

devnull

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Community · Accepted Answer · 2017-04-13 12:36:31Z

1

Try the eval command:

eval toprint='$'$dummy$i
echo $toprint

That's a nice one too!