bash, substitute part of variable by another variable

Question

#!/bin/bash

SUNDAY_MENU=BREAD
MONDAY_MENU=APPLES

TODAY=MONDAY


ECHO "I want ${${TODAY}_MENU}" # does not work, bad substitution

ECHO "I want ${`echo $TODAY`_MENU}" # does not work, bad substitution

Any Ideas ?

Answer 1

Use variable indirection like this:

varname=${TODAY}_MENU
echo ${!varname}

If you are using Bash 4 or later, however, you are probably better off using an associative array :

menu=([sunday]=bread [monday]=apples)
echo ${menu[$TODAY]}

Answer 2

I use eval function

#!/bin/bash

SUNDAY_MENU=BREAD
MONDAY_MENU=APPLES

TODAY=MONDAY

eval TODAY_MENU=\$\{${TODAY}_MENU\}

echo "I want ${TODAY_MENU}"