Pascal Pilz用100%純Bash重寫解決方案(沒有外部命令)：

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

例如,

join_by , a "b c" d #a,b c,d

join_by / var local tmp #var/local/tmp

join_by , "${FOO[@]}" #a,b,c

或者，我們可以使用printf來支持多字符分隔符，使用@gniourf_gniourf的思想。

function join_by { local d=$1; shift; echo -n "$1"; shift; printf "%s" "${@/#/$d}"; }

例如,

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

另一個解決辦法是：

#!/bin/bash

foo=('foo bar' 'foo baz' 'bar baz')

bar=$(printf ",%s" "${foo[@]}")

bar=${bar:1}

echo $bar

編輯：相同，但對于多字符可變長度分隔符：

#!/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

$ foo=(a "b c" d)$ bar=$(IFS=, ; echo "${foo[*]}")$ echo "$bar"a,b c,d