使用perl实现多字符分隔符:

function join {
   perl -e '$s = shift @ARGV; print join($s, @ARGV);' "$@"; 
}

join ', ' a b c # a, b, c

或者在一行中:

perl -le 'print join(shift, @ARGV);' ', ' 1 2 3
1, 2, 3

还有另一个解决方案:

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

echo $bar

编辑:相同，但用于多字符可变长度分隔符:

#!/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

这种方法处理值中的空格，但需要一个循环:

#!/bin/bash

FOO=( a b c )
BAR=""

for index in ${!FOO[*]}
do
    BAR="$BAR,${FOO[$index]}"
done
echo ${BAR:1}

也许迟到了，但这对我来说是可行的:

function joinArray() {
  local delimiter="${1}"
  local output="${2}"
  for param in ${@:3}; do
    output="${output}${delimiter}${param}"
  done

  echo "${output}"
}

如果你在循环中构建数组，这里有一个简单的方法:

arr=()
for x in $(some_cmd); do
   arr+=($x,)
done
arr[-1]=${arr[-1]%,}
echo ${arr[*]}

顶部答案的简短版本:

joinStrings() { local a=("${@:3}"); printf "%s" "$2${a[@]/#/$1}"; }

用法:

joinStrings "$myDelimiter" "${myArray[@]}"