Bash scripting - Iterating through "variable" variable names for a list of associative arrays

Question

I've got a variable list of associative arrays that I want to iterate through and retrieve their key/value pairs.

I iterate through a single associative array by listing all its keys and getting the values, ie.

for key in "${!queue1[@]}" do
    echo "key : $key"
    echo "value : ${queue1[$key]}"
done

The tricky part is that the names of the associative arrays are variable variables, e.g. given count = 5, the associative arrays would be named queue1, queue2, queue3, queue4, queue5.

I'm trying to replace the sequence above based on a count, but so far every combination of parentheses and eval has not yielded much more then bad substitution errors. e.g below:

for count in {1,2,3,4,5} do
    for key in "${!queue${count}[@]}" do
        echo "key : $key"
        echo "value : ${queue${count}[$key]}"
    done
done

Help would be very much appreciated!

The existing answers are largely out-of-date in the context of namevar support in bash 4.3 (and later). — Charles Duffy
– Charles Duffy, Commented Sep 29, 2016 at 20:52

Shawn Chin · Accepted Answer · 2012-11-09 11:06:32Z

3

The difficulty here stems from the fact that the syntax for indirect expansion (${!nameref}) clashes with the syntax for extracting keys from an associative arrays (${!array[@]}). We can have only one or the other, not both.

Wary as I am about using eval, I cannot see a way around using it to extract the keys of an indirectly referenced associative array:

keyref="queue${count}[@]"
for key in $(eval echo '${!'$keyref'}'); do ... ; done

You can however avoid eval and use indirect expansion when extracting a value from an array given the key. Do note that the [key] suffix must be part of the expansion:

valref="queue${count}[$key]"
echo ${!valref}

To put this in context:

for count in {1..5} ; do
    keyref="queue${count}[@]"
    for key in $(eval echo '${!'$keyref'}'); do
        valref="queue${count}[$key]"
        echo "key = $key"
        echo "value = ${!valref}"
    done
done

edited Nov 9, 2012 at 11:06

answered Nov 9, 2012 at 10:59

Shawn Chin

87.4k20 gold badges168 silver badges193 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

choroba · Accepted Answer · 2012-11-09 10:10:57Z

1

I was able to make it work with the following script:

for count in {1..5} ; do
    for key in $(eval echo '${!q'$count'[@]}') ; do
        eval echo '${q'$count"[$key]}"
    done
done

Note it breaks if any key contained a space. If you want to deal with complex data structures, use a more powerful language like Perl.

answered Nov 9, 2012 at 10:10

choroba

245k27 gold badges221 silver badges304 bronze badges

Comments

chepner · Accepted Answer · 2012-11-09 14:42:20Z

0

I think this might work (but untested). The key is to treat the indexing as the full name of a variable. (That is, the array queue5 can be treated as a sequence of variables named queue5[this], queue5[that], etc.)

for count in {1,2,3,4,5} do
    assoc="queue$count[@]"
    for key in "${!assoc}" do
        echo "key : $key"
        val="queue$count[$key]"
        echo "value : ${!val}"
    done
done

answered Nov 9, 2012 at 14:42

chepner

538k77 gold badges594 silver badges746 bronze badges

2 Comments

doubleDown Over a year ago

This seems to iterate through the values in the array (instead of the keys in the array). I guess you can dig out the key using the value with some work, but there might be a problem with non-unique values.

Shawn Chin Over a year ago

Indirect referencing works for accessing array values but not for keys. See my answer for details.

Collectives™ on Stack Overflow

Bash scripting - Iterating through "variable" variable names for a list of associative arrays

3 Answers 3

Comments

Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related