Assign to dynamic variable name in Bash

Question

How do I assign a value to variable that has a variable in its name?

var1="file"
var2_$var1="folder"

The code above gives me the error -bash: var2_file=folder: command not found. I was curious to know how to assign to a variable with another variable in its name.

Version of Bash is "GNU bash, version 4.1.2"

This is very shell specific. What shell are you using? Bash (from the error message)? Please list the version as well. Do not respond in a comment. Please edit the question itself. — Mad Physicist, Feb 09 '17 at 15:33
Why do you need it? There might be better ways to achieve your goal. — choroba, Feb 09 '17 at 15:45
[BashFAQ #6](http://mywiki.wooledge.org/BashFAQ/006) covers this in detail -- both associative arrays *and* indirect assignment as such. — Charles Duffy, Feb 09 '17 at 18:00

score 5 · Answer 1 · answered Feb 09 '17 at 15:35

5

With bash you can use declare:

declare var2_$var1="123"

answered Feb 09 '17 at 15:35

hek2mgl

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Would `export` work the same way? What about `set`? – Mad Physicist Feb 09 '17 at 15:37
Thanks alot for both of your answers. I tried `export`, `set` and `declare` all worked fine. But know how to retrieve value of it.!! `echo $var2_$var1` The output of it should be "123" – vardhan Feb 09 '17 at 15:44
Yes, that's the problem with this approach. I recomend to use and assoc array instead – hek2mgl Feb 09 '17 at 15:50

score 0 · Answer 2 · answered Feb 09 '17 at 17:16

How about using another variable to hold the dynamic name and use it for retrieving the value after setting?

new_var=var2_$var1
declare var2_$var1="123"
echo "${!new_var}" # => 123

Unfortunately, Bash doesn't allow declare $new_var="123" - that would have made this a little prettier.

Assign to dynamic variable name in Bash

2 Answers2