Variation on a Variable Variable in Bash

Question

I am looking for a way to add a string to a variable name in bash and evaluate this as a new variable. Example:

#!/bin/bash
file="filename"
declare ${file}_info="about a file"
declare ${file}_status="status of file"
declare ${file}_number="29083451"

echo ${${file}_info}  # <-- This fails with error: "bad substitution"

Is there a way to do this?

I'm not actually implementing this in any production code anywhere. I just want to know if there is some way to create a variable name using a variable and a string. It seemed like an interesting problem.

Also note that I am not asking about bash indirection, which is the use of ${!x} to evaluate the contents of a variable as a variable.

It looks like the eval approach given in stackoverflow.com/questions/8435256/… might be the one you are looking for. — jpe
– jpe, Commented Oct 6, 2014 at 21:54
The indirection FAQ tells you how to write, not just how to read, so it actually is the same topic. — Charles Duffy
– Charles Duffy, Commented Oct 6, 2014 at 22:09
Agreed. As I have reviewed comments and answers, it would seem that indirection must be involved in the solution. I was hoping for another way, but it seems that there is at least one solution if indirection is used in the process. — vastlysuperiorman
– vastlysuperiorman, Commented Oct 6, 2014 at 22:10
@jpe, re: eval, see BashFAQ #48: mywiki.wooledge.org/BashFAQ/048 — Charles Duffy
– Charles Duffy, Commented Oct 6, 2014 at 22:29

choroba · Accepted Answer · 2014-10-06 23:18:42Z

0

You aren't asking about indirection, but that's what can help you:

info=$file\_info
echo ${!info}

edited Oct 6, 2014 at 23:18

answered Oct 6, 2014 at 22:04

choroba

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1 Comment

vastlysuperiorman Over a year ago

Apparently my edit was rejected. It seems that your answer should say ${!info} instead of ${!name}. I had also hoped to achieve this without using a placeholder variable like "info" that you are using here, but you have indeed solved the problem. Will accept your answer.

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