How can I join elements of an array in Bash?

Question

If I have an array like this in Bash:

FOO=( a b c ) 

How do I join the elements with commas? For example, producing a,b,c.

Answer 1

A 100% pure Bash function that supports multi-character delimiters is:

function join_by {   local d=${1-} f=${2-}   if shift 2; then     printf %s "$f" "${@/#/$d}"   fi } 

For example,

join_by , a b c #a,b,c join_by ' , ' a b c #a , b , c join_by ')|(' a b c #a)|(b)|(c join_by ' %s ' a b c #a %s b %s c join_by $'\n' a b c #a<newline>b<newline>c join_by - a b c #a-b-c join_by '\' a b c #a\b\c join_by '-n' '-e' '-E' '-n' #-e-n-E-n-n join_by , # join_by , a #a 

The code above is based on the ideas by @gniourf_gniourf, @AdamKatz, @MattCowell, and @x-yuri. It works with options errexit (set -e) and nounset (set -u).

Alternatively, a simpler function that supports only a single character delimiter, would be:

function join_by { local IFS="$1"; shift; echo "$*"; } 

For example,

join_by , a "b c" d #a,b c,d join_by / var local tmp #var/local/tmp join_by , "${FOO[@]}" #a,b,c 

This solution is based on Pascal Pilz's original suggestion.

A detailed explanation of the solutions previously proposed here can be found in "How to join() array elements in a bash script", an article by meleu at dev.to.

Answer 2

Yet another solution:

#!/bin/bash foo=('foo bar' 'foo baz' 'bar baz') bar=$(printf ",%s" "${foo[@]}") bar=${bar:1}  echo $bar 

Edit: same but for multi-character variable length separator:

#!/bin/bash separator=")|(" # e.g. constructing regex, pray it does not contain %s foo=('foo bar' 'foo baz' 'bar baz') regex="$( printf "${separator}%s" "${foo[@]}" )" regex="${regex:${#separator}}" # remove leading separator echo "${regex}" # Prints: foo bar)|(foo baz)|(bar baz