Is it possible to print the content of the content of a variable with shell script? (indirect referencing)
You can accomplish this using bash's indirect variable expansion (as long as it's okay for you to leave out the $
from your reference variable):
$ var=test
$ test="my string"
$ echo "$var"
test
$ echo "${!var}"
my string
3.5.3 Shell Parameter Expansion
For the case when the variable name contained in var
is prefixed with $
you may use eval
:
$ var='$test'
$ test="my string"
$ eval echo $var
my string
What happens here:
- bash expands
$var
to the value$test
, producing aeval echo $test
command; eval
evaluatesecho $test
expression and produces a desired result.
Note, that using eval
in general may be dangerous (depending on what is stored in var
), so prefer avoiding it. Indirect expansion feature is better for your case (but you need to get rid of $
sign in $test
).
Similar to Jesse_b's answer, but using a name reference variable instead of variable indirection (requires bash
4.3+):
$ declare -n var=test
$ test="my string"
$ echo "$var"
my string
The name reference variable var
holds the name of the variable it's referencing. When the variable is dereferenced as $var
, that other variable's value is returned.
bash
resolves name references recursively:
$ declare -n var1=var2
$ declare -n var2=test
$ test="hello world"
$ echo "$var1"
hello world
For completeness, using an associative array (in bash
4.0+) is also one way of solving this, depending on requirements:
$ declare -A strings
$ strings[test]="my string"
$ var=test
$ echo "${strings[$var]}"
my string
This provides a more flexible way of accessing more than one value by a key or name that may be determined dynamically. This may be preferable if you want to collect all values of a particular category in a single array, but still be able to access them by some key (e.g. names accessible by ID, or pathnames accessible by purpose etc.) as it does not pollute the variable namespace of the script.