Unable to print value of shell variable stored in another variable [duplicate]

Question

I'm trying to find all variables which match a particular pattern and print their values.

test_a="apple"
test_b="banana"
test_c="carrot"
test_d="doughnut"

test_show_all () {
    local i
    for i in ${!test_*}; do
        printf "..$i\n"
    #   printf "..$i-->${$i}\n"
    done    
}

The loop is finding the correct variables. But if I uncomment the second line of the for loop, bash is unhappy with the syntax ${$i}. I thought this should work since $i holds the name of a variable, so I thought ${$i} should expand to value of that stored name.

user000001 · Accepted Answer · 2014-06-06 16:44:27Z

5

The indirect variable reference is ${!var}. Change your code to

printf "..$i-->${!i}\n"

and it should work.

answered Jun 6, 2014 at 16:44

user000001

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1 Comment

Aleks-Daniel Jakimenko-A. Over a year ago

Bash pitfall #32

chepner · Accepted Answer · 2014-06-06 17:01:14Z

1

Not useful now, but in bash 4.3, you will also be able to use namerefs.

for name in ${!test_*}; do
  declare -n value=$name
  printf "..$name->$value\n"
done

As a short cut, if you only want to iterate over the values (without regard to the actual name of the variable), you can use

declare -n value
for value in ${!test_*}; do
  printf "..$value\n"
done

answered Jun 6, 2014 at 17:01

chepner

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