Using variable as part of another variable name

Question

I'm trying to use a variable as part of another variable's name.

For example, I have some variables in a CONFIG_ scope.

echo $CONFIG_systema_value > "testconfig"
echo $CONFIG_systemb_value > "testconfig2"

and then I have a "deciding" variable like WANTEDSYSTEM

echo $WANTEDSYSTEM > "systema"

I need to be able to "dynamically" switch the config depending on the wantedsystem variable.

My initial attempts I tried something like this echo $CONFIG_$WANTEDSYSTEM_value which I would like to have output testconfig as if I was asking it to echo $CONFIG_systema_value

larsks · Accepted Answer · 2020-11-10 14:32:36Z

5

You can use the ${!name} syntax to indirectly reference a variable, like this:

CONFIG_systema_value=foo
CONFIG_systemb_value=bar
WANTEDSYSTEM=systema

target_var_name="CONFIG_${WANTEDSYSTEM}_value"
echo ${!target_var_name}

(Note that this is bash-specific syntax, but it sounds like that's what you're looking for.)

answered Nov 10, 2020 at 14:32

larsks

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2 Comments

l00p Over a year ago

Isn't it any trick with eval to make it in one line with echo?

larsks Over a year ago

Sure, you could write something like eval echo "\$CONFIG_${WANTEDSYSTEM}_value", but I avoid eval when possible because if you're not careful what you pass to it it can have unintended side effects.

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