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
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 $b
1,a b,3
其他回答
我相信这是最短的解决方案,正如Benamin W.已经提到的:
(IFS=,; printf %s "${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
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}
$ set a 'b c' d
$ history -p "$@" | paste -sd,
a,b c,d
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
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}"
}