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。
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
$ set a 'b c' d
$ history -p "$@" | paste -sd,
a,b c,d
其他回答
awk -v sep=. 'BEGIN{ORS=OFS="";for(i=1;i<ARGC;i++){print ARGV[i],ARGC-i-1?sep:""}}' "${arr[@]}"
or
$ a=(1 "a b" 3)
$ b=$(IFS=, ; echo "${a[*]}")
$ echo $b
1,a b,3
我相信这是最短的解决方案,正如Benamin W.已经提到的:
(IFS=,; printf %s "${a[*]}")
想要添加,如果你使用zsh,你可以删除子shell:
IFS=, printf %s "${a[*]}"
测试:
a=(1 'a b' 3)
IFS=, printf %s "${a[*]}"
1,a b,3
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 $
$
下面是一个100%纯Bash函数,它可以完成这项工作:
join() {
# $1 is return variable name
# $2 is sep
# $3... are the elements to join
local retname=$1 sep=$2 ret=$3
shift 3 || shift $(($#))
printf -v "$retname" "%s" "$ret${@/#/$sep}"
}
看:
$ a=( one two "three three" four five )
$ join joineda " and " "${a[@]}"
$ echo "$joineda"
one and two and three three and four and five
$ join joinedb randomsep "only one element"
$ echo "$joinedb"
only one element
$ join joinedc randomsep
$ echo "$joinedc"
$ a=( $' stuff with\nnewlines\n' $'and trailing newlines\n\n' )
$ join joineda $'a sep with\nnewlines\n' "${a[@]}"
$ echo "$joineda"
stuff with
newlines
a sep with
newlines
and trailing newlines
$
这甚至保留了末尾的换行符,并且不需要一个子shell来获得函数的结果。如果你不喜欢printf -v(为什么你不喜欢它?)并传递一个变量名,你当然可以为返回的字符串使用一个全局变量:
join() {
# $1 is sep
# $2... are the elements to join
# return is in global variable join_ret
local sep=$1 IFS=
join_ret=$2
shift 2 || shift $(($#))
join_ret+="${*/#/$sep}"
}
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[*]}