How to join arguments with a separator?

Question

In bash, how to join arguments of a function into a single string?

The seperator is fixed, and is not space (thus "$*" is not what I want). Here use ", " as example.

join_args() {
# join args with sep ", ", then echo the joined string
# ... code here
}

join_args abc def ghi
# abc, def, ghi

choroba · Accepted Answer · 2018-07-17T05:56:59.070

2

$* can work with characters other than space if you set $IFS, but it can only use one character. Thus

join_args() {
    local IFS=', '
    echo "${*}"
}

outputs abc,def,ghi.

If you need longer separators, you have to use a loop:

join_args() {
    while (($# > 1)) ; do
        printf '%s, ' "$1"
        shift
    done
    if (($#)) ; then
        printf '%s\n' "$1"
    fi
}

Or, use a real programming language:

perl -E 'say join ", ", @ARGV' abc def ghi

edited Jul 17 '18 at 05:56

answered Jul 17 '18 at 05:28

choroba

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Your second `join_args()` will loose arg if the args count > 3 – Konstantin Vustin Jul 17 '18 at 05:38
@KonstantinVustin: Oops, thanks. Fixed. – choroba Jul 17 '18 at 05:57

dawg · Answer 2 · 2018-07-17T05:47:21.147

You can try:

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

Then use by:

$ join_by , abc def ghi
abc,def,ghi

If you want the extra space after the ", ", use sed I suppose:

$ join_by , abc def ghi | sed -e 's/,/, /g'
abc, def, ghi

Or you can use printf with parameter expansion thus:

$ function join_by { local d=$1; shift; echo -n "$1"; shift; printf "%s" "${@/#/$d}"; }
$ arr=( abc def ghi "kl mn op" )
$ echo $(join_by ", " "${arr[@]}")
abc, def, ghi, kl mn op

score 0 · Answer 3 · answered Jul 17 '18 at 05:30

#!/usr/bin/env bash

join_args() {
  count=$#
  for i in `seq 1 $count`
  do
   if [[ $i = $count ]]
   then
     joined+=("$1")
   else
    joined+=("$1,")
    shift
   fi
  done
echo ${joined[*]}
}

join_args abc def ghi sfl ugw pwg afm

Output:

abc, def, ghi, sfl, ugw, pwg, afm

How to join arguments with a separator?

3 Answers3

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