using variable to make variable name in shell script

Question

I want to echo $ANMIAL0 and then $ANIMAL1 using a script below. But I get line 7: ${ANIMAL$i}: bad substitution error message. What's wrong?

#!/bin/sh
ANIMAL0="tiger"
ANIMAL1="lion"
i=0
while test $i -lt 2; do
echo "Hey $i !"
echo ${ANIMAL$i}
i=`expr $i + 1`
done

Semirix · Accepted Answer · 2014-06-23 11:30:53Z

1

You're probably better off using an array instead of ANIMAL0 and ANIMAL1. Something like this maybe?

#!/bin/bash

animals=("Tiger" "Lion")

for animal in ${animals[*]}
do
    printf "Hey, ${animal} \n"
done

Using eval will get you into trouble down the road and is not best practice for what you're trying to do.

edited Jun 23, 2014 at 11:30

answered Jun 23, 2014 at 11:22

Semirix

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1 Comment

Semirix Over a year ago

Well, if you do end up using this method instead please mark this as the correct answer. Other than that, glad to help :)

tripleee · Accepted Answer · 2014-06-23 11:21:33Z

0

The problem is that the shell normally performs a single substitution pass. You can force a second pass with eval, but this obviously comes with the usual security caveats (don't evaluate unchecked user input, etc).

eval echo \$ANIMAL$i

Bash has various constructs to help avoid eval.

answered Jun 23, 2014 at 11:21

tripleee

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Comments

Kalanidhi · Accepted Answer · 2014-06-23 11:19:12Z

0

You can use eval

eval echo \$ANIMAL$i

answered Jun 23, 2014 at 11:19

Kalanidhi

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