Create a variable from another one in bash

Question

I have a variable $var. I want to create another variable named test_$var.

How can I call this variable after the initialization?
echo $test_$var doesn't work.

When you say `test_$var` you mean `test_` + `content of $var` or just `test_var`? — fedorqui, Sep 15 '14 at 10:48
Related: [Is it possible to build variable names from other variables in bash?](http://stackoverflow.com/q/3963494/1983854) — fedorqui, Sep 15 '14 at 10:53

score 1 · Accepted Answer · answered Sep 15 '14 at 10:48

1

You can use declare shell bulletin:

var='abc'
var1="$test_$var"
declare $var1=5
echo "${!var1}"
5

answered Sep 15 '14 at 10:48

anubhava

The initialization works for me but I want to echo "$test_$var", not "$test_abc" – Ciprian Vintea Sep 15 '14 at 10:50
I ran into this situation lately myself and all the answers on this and related questions always seem to use declare. Is there a reason to prefer declare over eval? In my case I had to use eval since my shell is busybox and does not know declare. (for eval examples, have a look at [tldp](http://www.tldp.org/LDP/abs/html/ivr.html)) – a.peganz Sep 24 '14 at 10:34
`eval` is very dangerous unless you've 100% guarantee of controlling all the text that is used in `eval`. – anubhava Sep 24 '14 at 10:47

1 Answers1