Bash - variable variables

Question

I have the variable $foo="something" and would like to use:

bar="foo"; echo $($bar)

to get "something" echoed.

Please see [BashFAQ/006](http://mywiki.wooledge.org/BashFAQ/006). Also, you shouldn't try to use a dollar sign on the left side of an assignment. — Dennis Williamson, May 25 '12 at 15:56

score 108 · Accepted Answer · edited Dec 18 '20 at 05:44

108

In bash, you can use ${!variable} to use variable variables.

foo="something"
bar="foo"
echo "${!bar}"

# something

edited Dec 18 '20 at 05:44

Romain

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answered May 25 '12 at 15:49

dAm2K

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3

in `sh` it says bad substitution. Any idea how to do it in `sh`? – Shiplu Mokaddim Sep 06 '13 at 06:00
how does this work with arrays? – Edison Apr 14 '14 at 21:14
1

@Edison `foo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]} ` – dAm2K Apr 14 '14 at 23:23
I ended up doing something like " read -a subNames <<< $(eval "echo \${$rootName[@]}") for subName in "${subNames[@]}"" – Edison Apr 15 '14 at 01:42
@Edison see also [my recent answer](http://stackoverflow.com/a/25167219/2908724). – bishop Aug 06 '14 at 18:13
What do if bar somting like bar=$(${LINEDATA[3]}|iconv -f "windows-1251" -t "UTF-8") ? – Andrey Jul 31 '17 at 19:14

score 9 · Answer 2 · edited Mar 26 '16 at 13:16

9

eval echo \"\$$bar\" would do it.

edited Mar 26 '16 at 13:16

Gilles 'SO- stop being evil'

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answered May 25 '12 at 15:44

Matt K

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Be aware of the [security implications of `eval`](http://mywiki.wooledge.org/BashFAQ/048). – Dennis Williamson May 25 '12 at 15:58
3

This solution has the benefit of being POSIX-compatible for non-Bash shells (for example, lightweight environments like embedded systems or Docker containers). And you can assign the value to another variable like so: ```sh var=$(eval echo \"\$$bar\") ``` – Jason Suárez Feb 03 '17 at 04:29

score 9 · Answer 3 · answered Aug 06 '14 at 18:10

9

The accepted answer is great. However, @Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[@]", so that the array is expanded with the "!". Check out this function to dump variables:

$ function dump_variables() {
    for var in "$@"; do
        echo "$var=${!var}"
    done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[@]

This outputs:

STRING=Hello World
ARRAY=ab
ARRAY[@]=ab cd

When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[@] format, you get the array and all its values expanded.

answered Aug 06 '14 at 18:10

bishop

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Good point about handling arrays. Any idea how to get the *indices* of an array? The [manual](http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion) indicates this is normally done with `${!ARRAY[@]}`, which seems to conflict with the variable indirection syntax. – dimo414 Aug 28 '14 at 05:33
@dimo414 Yeah, getting the keys through indirection is trickier. You'd have to pass just the name, then do the expansion in the method: `local -a 'keys=("${!'"$var"'[@]}")'`. The [indirection article on Bash Hackers](http://wiki.bash-hackers.org/syntax/arrays#indirection) goes into more depth. – bishop Aug 28 '14 at 13:20

score 1 · Answer 4 · answered May 16 '16 at 05:56

1

To make it more clear how to do it with arrays:

arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation 
# of the variable (array) as its value:
var=arr[@]         
echo "${!var}"

answered May 16 '16 at 05:56

Jahid

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Bash - variable variables

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