I have the variable $foo="something" and would like to use:
bar="foo"; echo $($bar)
to get "something" echoed.
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3Please see BashFAQ/006. Also, you shouldn't try to use a dollar sign on the left side of an assignment.Dennis Williamson– Dennis Williamson2012-05-25 15:56:42 +00:00Commented May 25, 2012 at 15:56
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4 Answers
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
# something
7 Comments
Shiplu Mokaddim
in
sh it says bad substitution. Any idea how to do it in sh?
Edison
how does this work with arrays?
dAm2K
@Edison
foo1="something1" foo2="something2" bar[0]="foo1" bar[1]="foo2" echo ${!bar[0]} echo ${!bar[1]} Edison
I ended up doing something like " read -a subNames <<< $(eval "echo \${$rootName[@]}") for subName in "${subNames[@]}""
bishop
@Edison see also my recent answer.
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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.
2 Comments
bishop
@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 goes into more depth.eval echo \"\$$bar\" would do it.
2 Comments
Dennis Williamson
Be aware of the security implications of
eval.
Jason Suárez
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\") 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}"
