如果我在Bash中有一个这样的数组:

FOO=( a b c )

如何用逗号连接元素?例如,生成a b c。


当前回答

顶部答案的简短版本:

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

用法:

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

其他回答

我将数组作为字符串回显,然后将空格转换为换行,然后使用粘贴将所有内容连接到一行,如下所示:

tr " " "\n" <<< "$FOO" |粘贴-sd, -

结果:

a, b, c

对我来说,这似乎是最快最干净的!

我的尝试。

$ array=(one two "three four" five)
$ echo "${array[0]}$(printf " SEP %s" "${array[@]:1}")"
one SEP two SEP three four SEP five

不使用外部命令:

$ FOO=( a b c )     # initialize the array
$ BAR=${FOO[@]}     # create a space delimited string from array
$ BAZ=${BAR// /,}   # use parameter expansion to substitute spaces with comma
$ echo $BAZ
a,b,c

警告,它假设元素没有空格。

这个特别适用于busybox的sh和$@:

$ FOO=(a b c)
$ printf '%s\n' "${FOO[@]}" | paste -sd,
a,b,c

Or:

join_by() {
    local d=$1
    shift
    printf '%s\n' "$@" | paste -sd "$d"
}
join_by , "${FOO[@]}"  # a,b,c

感谢@gniourf_gniourf对我迄今为止的最佳世界组合的详细评论。很抱歉发布的代码没有完全设计和测试。这是一个更好的尝试。

# join with separator
join_ws() { local d=$1 s=$2; shift 2 && printf %s "$s${@/#/$d}"; }

这种概念上的美是

(still) 100% pure bash ( thanks for explicitly pointing out that printf is a builtin as well. I wasn't aware about this before ... ) works with multi-character delimiters more compact and more complete and this time carefully thought over and long-term stress-tested with random substrings from shell scripts amongst others, covering use of shell special characters or control characters or no characters in both separator and / or parameters, and edge cases, and corner cases and other quibbles like no arguments at all. That doesn't guarantee there is no more bug, but it will be a little harder challenge to find one. BTW, even the currently top voted answers and related suffer from such things like that -e bug ...

附加的例子:

$ join_ws '' a b c
abc
$ join_ws ':' {1,7}{A..C}
1A:1B:1C:7A:7B:7C
$ join_ws -e -e
-e
$ join_ws $'\033[F' $'\n\n\n'  1.  2.  3.  $'\n\n\n\n'
3.
2.
1.
$ join_ws $ 
$