Array of arrays in bash

Question

I\'m attempting to read an input file line by line which contains fields delimited by periods. I want to put them into an array of arrays so I can loop through them later on

Answer 1

I struggled with this but found an uncomfortable compromise. In general, when faced with a problem whose solution involves using data structures in Bash, you should switch to another language like Python. Ignoring that advice and moving right along:

My use cases usually involve lists of lists (or arrays of arrays) and looping over them. You usually don't want to nest much deeper than that. Also, most of the arrays are strings that may or may not contain spaces, but usually don't contain special characters. This allows me to use not-to-confusing syntax to express the outer array and then use normal bash processing on the strings to get a second list or array. You will need to pay attention to your IFS delimiter, obvi.

Thus, associative arrays can give me a way to create a list of lists like:

declare -A JOB_LIST=(
   [job1] = "a set of arguments"
   [job2] = "another different list"
   ...
)

This allows you to iterate over both arrays like:

for job in "${!JOB_LIST[@]}"; do
  /bin/jobrun ${job[@]}
done

Ah, except that the output of the keys list (using the magical ${!...}) means that you will not traverse your list in order. Therefore, one more necessary hack is to sort the order of the keys, if that is important to you. The sort order is up to you; I find it convenient to use alphanumerical sorting and resorting to aajob1 bbjob3 ccjob6 is perfectly acceptable.

Therefore

declare -A JOB_LIST=(
   [aajob1] = "a set of arguments"
   [bbjob2] = "another different list"
   ...
)
sorted=($(printf '%s\n' "${!JOB_LIST[@]}"| /bin/sort))
for job in "${sorted[@]}"; do
   for args in "${job[@]}"; do
     echo "Do something with ${arg} in ${job}"
   done
done