How do I join an array of strings where each string has spaces?

Question

My bash script:

#!bin/bash
MY_ARRAY=("Some string" "Another string")
function join { local IFS="$1"; shift; echo -e "$*"; }
join "," ${MY_ARRAY[@]}

I want the output to be: Some string,Another string.

Instead, I get Some,string,Another,string.

What must I change to get the result I want?

@jasonwryan I corrected a couple typos in my question. Also, is there a solution for when I want to do this for a longer array? — Username, Aug 30 '17 at 22:26
Add double quotes around ${MY_ARRAY[@]}". See also Why does my shell script choke on whitespace or other special characters? — Wildcard, Aug 30 '17 at 22:26
And Are there naming conventions for variables in shell scripts? — Wildcard, Aug 30 '17 at 22:28
Clarifying question: what behavior do you want if the array is empty? — Wildcard, Aug 30 '17 at 22:34
@Username, okay, so if it's empty, the function should output a single newline? — Wildcard, Aug 30 '17 at 22:43

score 10 · Accepted Answer · answered Aug 30 '17 at 22:47

My modified version of your script:

#!bin/bash
my_array=("Some string" "Another string")
my_join() {
  [ "$#" -ge 1 ] || return 1
  local IFS="$1"
  shift
  printf '%s\n' "$*"
}
my_join , "${my_array[@]}"

Notes:

Don't use uppercase variable names
- See Are there naming conventions for variables in shell scripts?
join is a standard tool. Use a different name.
Use quoting
- See Why does my shell script choke on whitespace or other special characters?
Use printf, not echo
- See Why is printf better than echo?

How do I join an array of strings where each string has spaces?

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