Accessing bash associative array using variables

Question

I have spent a few hours on this now and can't seem to find a solution.

I have set three associative arrays in code:

template1 template2 template3

This could really be n number of arrays and a counter variable is also set in code so that:

template_name="template$counter"

Would give the next template to create. So my issue is when I want to loop over n number of associative arrays, I need to do something like:

for (( i = 1; i < $counter; i++ )); do
  template_name="template$i"
    for i in ${!template_name[@]}; do
      echo "$i - ${template_name[$i]}"
    done
done

But it does not work - I get bad substitution errors. It seems that I can;t use a variable name to access associative arrays dynamically. Is there any other way to look at this problems ?

You should either: (1) rethink your design or (2) change language. Doing this stuff in bash is not really advised. — gniourf_gniourf
– gniourf_gniourf, Commented Jun 26, 2013 at 7:00

ilium007 · Accepted Answer · 2013-06-26 03:52:43Z

1

I found the answer after some more digging:

eval echo "APP_NAME $template_name: "\${${template_name}[APP_NAME]}

answered Jun 26, 2013 at 3:52

ilium007

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ilium007 Over a year ago

I know - but I am stuck with bash here at work and can't find another method to achieve the outcome I need.

gxtaillon · Accepted Answer · 2013-10-07 23:47:27Z

0

I'm not exactly sure if this is want you expected but I think it can help you find a better solution than using eval.

#!/bin/bash
template0=( "123" "456" )
template1=( "zxc" "edc" )

counter=2
for (( i = 0; i != $counter; ++i )); do
    template_name="template$i[@]"
    echo ${!template_name}
    for j in ${!template_name}; do
        echo "$j - template$i"
    done
done

Produces:

==> 123 456
==> 123 - template0
==> 456 - template0
==> zxc edc
==> zxc - template1
==> edc - template1

answered Oct 7, 2013 at 23:47

gxtaillon

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