How to use a variable's value as another variable's name in bash

Question

I want to declare a variable, the name of which comes from the value of another variable, and I wrote the following piece of code:

a="bbb"
$a="ccc"

but it didn't work. What's the right way to get this job done?

If you are on Bash 5+ the better solution is nearly always an associative array. The duplicate has several demonstrations of this. — tripleee, Jan 04 '23 at 07:18

Kaz · Answer 1 · 2014-04-14T23:36:21.633

eval is used for this, but if you do it naively, there are going to be nasty escaping issues. This sort of thing is generally safe:

name_of_variable=abc

eval $name_of_variable="simpleword"   # abc set to simpleword

This breaks:

eval $name_of_variable="word splitting occurs"

The fix:

eval $name_of_variable="\"word splitting occurs\""  # not anymore

The ultimate fix: put the text you want to assign into a variable. Let's call it safevariable. Then you can do this:

eval $name_of_variable=\$safevariable  # note escaped dollar sign

Escaping the dollar sign solves all escape issues. The dollar sign survives verbatim into the eval function, which will effectively perform this:

eval 'abc=$safevariable' # dollar sign now comes to life inside eval!

And of course this assignment is immune to everything. safevariable can contain *, spaces, $, etc. (The caveat being that we're assuming name_of_variable contains nothing but a valid variable name, and one we are free to use: not something special.)

So the `$safevariable` is the `$safe_target_value` to assign the variable name saved in `$name_of_variable` ? — inetphantom, Apr 06 '18 at 12:54
`eval $name_of_variable="word\ splitting\ occurs"` also works — Stony, Aug 10 '21 at 01:35

score 62 · Answer 2 · answered May 06 '15 at 10:20

62

You can use declare and !, like this:

John="nice guy"
programmer=John
echo ${!programmer} # echos nice guy

Second example:

programmer=Ines
declare $programmer="nice gal"
echo $Ines # echos nice gal

answered May 06 '15 at 10:20

Flimm

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Hi Flimm. Is it possible with arrays ? Something like `echo "${!array[1]}"` ? – Jul 19 '17 at 18:42
@leaf: Maybe the `-a`-option of `declare` helps? See `declare --help`, and test for yourself. – Golar Ramblar May 22 '18 at 09:43
It is nice gesture to say such nice things about your colleagues :D:D – Mr. Developerdude Sep 26 '18 at 08:16
2

Just a note, this will not work inside a function, for that I found KAz' eval solution to work. – Mr. Developerdude Sep 26 '18 at 08:50
4

Bash docs call the `${!var_name}` form "indirect expansion": https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html – Michael Burr Feb 28 '20 at 18:26
@Mr.Developerdude Confirmed this approach works inside functions with bash `3.2.57(1)-release` – raffian May 28 '23 at 21:29

score 19 · Answer 3 · answered Mar 15 '12 at 07:51

19

This might work for you:

foo=bar
declare $foo=baz
echo $bar
baz

or this:

foo=bar
read $foo <<<"baz"
echo $bar
baz

answered Mar 15 '12 at 07:51

potong

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correct, better to use declare over eval (security reasons) – meso_2600 Apr 10 '15 at 13:48
@meso_2600 declare does not behave the same way as assignment, because it will change the scope of the variable to the local one (you won't be able to modify a variable from a parent scope with declare, unless you declare it as a global variable). – niieani Jan 12 '16 at 22:32
your second answer is great. no subshell spawned. cheers. – bhu Boue vidya Jul 31 '17 at 04:30
1

@niieani: According to `declare --help`, the `-g` option to `declare` always declares a global variable. – Golar Ramblar May 22 '18 at 09:42
@GolarRamblar that's correct. I didn't state otherwise, unless you're referring to something else? – niieani May 30 '18 at 22:01
I really like the `read` idea, cheers mate – leetbacoon Nov 28 '19 at 20:41

score 10 · Answer 4 · answered Mar 15 '12 at 06:37

10

You could make use of eval for this.
Example:

$ a="bbb"
$ eval $a="ccc"
$ echo $bbb
ccc

Hope this helps!

answered Mar 15 '12 at 06:37

another.anon.coward

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score 6 · Answer 5 · answered Oct 28 '13 at 12:35

6

If you want to get the value of the variable instead of setting it you can do this

var_name1="var_name2"
var_name2=value_you_want
eval temp_var=\$$var_name1
echo "$temp_var"

You can read about it here indirect references.

answered Oct 28 '13 at 12:35

bat_ventzi

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score -3 · Answer 6 · answered Mar 15 '12 at 06:36

You can assign a value to a variable using simple assignment using a value from another variable like so:

#!/usr/bin/bash

#variable one
a="one"

echo "Variable a is $a"
#variable two with a's variable
b="$a"

echo "Variable b is $b"

#change a
a="two"
echo "Variable a is $a"
echo "Variable b is $b"

The output of that is this:

Variable a is one
Variable b is one
Variable a is two
Variable b is one

So just be sure to assign it like this b="$a" and you should be good.

The question is how to indirect upon computed variable names, now how to assign expressions fixed variables. — Kaz, Mar 15 '12 at 07:07

How to use a variable's value as another variable's name in bash

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