Assign to dynamic variable name in Bash [duplicate]

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
– Mad Physicist, Commented Feb 9, 2017 at 15:33
Why do you need it? There might be better ways to achieve your goal. — choroba
– choroba, Commented Feb 9, 2017 at 15:45
BashFAQ #6 covers this in detail -- both associative arrays and indirect assignment as such. — Charles Duffy
– Charles Duffy, Commented Feb 9, 2017 at 18:00

hek2mgl · Accepted Answer · 2017-02-09 15:35:58Z

5

With bash you can use declare:

declare var2_$var1="123"

answered Feb 9, 2017 at 15:35

hek2mgl

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3 Comments

Mad Physicist Over a year ago

Would export work the same way? What about set?

vardhan Over a year ago

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"

hek2mgl Over a year ago

Yes, that's the problem with this approach. I recomend to use and assoc array instead

codeforester · Accepted Answer · 2017-02-09 17:16:21Z

0

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.

answered Feb 9, 2017 at 17:16

codeforester

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