in POSIX shell, how to print the value of the value of a variable

Question

In a POSIX shell:

>VALUE=value
>FOOBAR=VALUE

Now how to print the value of $FOOBAR? $$FOOBAR does not work - it goes left-to-right so interprets $$ first as process ID.

The duplicate question before me that people are pointing to, yes it does contain the answer to my question but it buried in a ton of irrelevant gunk. My question is much simpler and to the point.

@Bharel of course does not work, for the same reason as above - the shell interprets left to right — Mark Galeck, Mar 26 '16 at 13:05
http://stackoverflow.com/questions/10757380/bash-variable-variables — Karoly Horvath, Mar 26 '16 at 13:06
@KarolyHorvath yes there is the answer there in your link, use `eval`: `eval echo \$$FOOBAR` . Please make it an answer, I will accept. — Mark Galeck, Mar 26 '16 at 13:13
Possible duplicate of [Indirect parameter substitution in shell script](http://stackoverflow.com/questions/4380478/indirect-parameter-substitution-in-shell-script) — Gilles 'SO- stop being evil', Mar 26 '16 at 13:18

score 1 · Answer 1 · answered Mar 26 '16 at 13:17

For POSIX portability, you are pretty much left with

eval echo \$$variable

The usual caveats apply; don't use eval if the string you pass in is not completely and exclusively under your own control.

I lowercased your example; you should not be using upper case for your private variables, as uppercase names are reserved for system use.

score 0 · Accepted Answer · answered Mar 26 '16 at 13:16

The only way to do this in POSIX is to use eval, so be very certain that FOOBAR contains nothing more than a valid variable name.

$ VALUE=value
$ FOOBAR=VALUE
$ valid_var='[[:alpha:]_][[:alnum:]_]*$'
$ expr "$FOOBAR" : "$valid_var" > /dev/null && eval "echo \$$FOOBAR"

in POSIX shell, how to print the value of the value of a variable

2 Answers2