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 a eval echo $test command;
• eval evaluates echo $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

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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.