string 将 Bash 数组转换为分隔字符串
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Converting a Bash array into a delimited string
提问by koola
I would like to know the following;
我想知道以下内容；
- Why the given non-working example doesn't work.
- If there are any other cleaner methods than those given in working example.
- 为什么给定的非工作示例不起作用。
- 如果有任何其他比工作示例中给出的更清洁的方法。
Non-working example
非工作示例
> ids=(1 2 3 4);echo ${ids[*]// /|}
1 2 3 4
> ids=(1 2 3 4);echo ${${ids[*]}// /|}
-bash: ${${ids[*]}// /|}: bad substitution
> ids=(1 2 3 4);echo ${"${ids[*]}"// /|}
-bash: ${"${ids[*]}"// /|}: bad substitution
Working example
工作示例
> ids=(1 2 3 4);id="${ids[@]}";echo ${id// /|}
1|2|3|4
> ids=(1 2 3 4); lst=$( IFS='|'; echo "${ids[*]}" ); echo $lst
1|2|3|4
In context, the delimited string to be used in a sedcommand for further parsing.
在上下文中，要在sed命令中用于进一步解析的分隔字符串。
采纳答案by F. Hauri
# REVISION: 2017-03-14
# Use of read and other bash specific features (bashisms)
Because parentheses are used to delimit an array, not a string:
因为括号用于分隔数组，而不是字符串：
ids="1 2 3 4";echo ${ids// /|}
1|2|3|4
Some samples: Populating $idswith two strings: a band c d
一些示例：填充$ids两个字符串： a b和c d
ids=("a b" "c d")
echo ${ids[*]// /|}
a|b c|d
IFS='|';echo "${ids[*]}";IFS=$' \t\n'
a b|c d
... and finally:
......最后：
IFS='|';echo "${ids[*]// /|}";IFS=$' \t\n'
a|b|c|d
Where array is assembled, separated by 1st char of $IFS, but with space replaced by |in each element of array.
组装数组的地方，由 的第一个字符分隔$IFS，但|在数组的每个元素中替换为空格。
When you do:
当你这样做时：
id="${ids[@]}"
you transfer the string build from the merging of the arrayidsby a space to a new variable of type string.
您从的合并传送串构建阵列ids通过的空间类型的新变量字符串。
Note:when "${ids[@]}"give a space-separatedstring, "${ids[*]}"(with a star *instead of the at sign @) will render a string separated by the first character of $IFS.
注意：当"${ids[@]}"给出一个空格分隔的字符串时，"${ids[*]}"（用星号*而不是 at 符号@）将呈现一个由 的第$IFS一个字符分隔的字符串。
what man bashsays:
什么man bash说：
man -Len -Pcol\ -b bash | sed -ne '/^ *IFS /{N;N;p;q}'
IFS The Internal Field Separator that is used for word splitting
after expansion and to split lines into words with the read
builtin command. The default value is ``<space><tab><newline>''.
Playing with $IFS:
玩$IFS：
set | grep ^IFS=
IFS=$' \t\n'
set | grep ^IFS=
IFS=$' \t\n'
declare -p IFS
declare -- IFS="
"
printf "%q\n" "$IFS"
$' \t\n'
Literally a space, a tabulationand (meaning or)a line-feed. So, while the first character is a space. the use of *will do the same as @.
字面意思是 a space、 atabulation和（意思是或）a line-feed。所以，虽然第一个字符是一个空格。的使用*与@.
But:
但是：
{
# OIFS="$IFS"
# IFS=$': \t\n'
# unset array
# declare -a array=($(echo root:x:0:0:root:/root:/bin/bash))
# OIFS="$IFS"
# IFS=$': \t\n'
# unset array
# declare -a array=($(echo root:x:0:0:root:/root:/bin/bash))
IFS=: read -a array < <(echo root:x:0:0:root:/root:/bin/bash)
echo 1 "${array[@]}"
echo 2 "${array[*]}"
OIFS="$IFS" IFS=:
echo 3 "${array[@]}"
echo 4 "${array[*]}"
IFS="$OIFS"
}
1 root x 0 0 root /root /bin/bash
2 root x 0 0 root /root /bin/bash
3 root x 0 0 root /root /bin/bash
4 root:x:0:0:root:/root:/bin/bash
Note:The line IFS=: read -a array < <(...)will use :as separator, without setting $IFSpermanently. This is because output line #2present spaces as separators.
注：该生产线IFS=: read -a array < <(...)将使用:作为分隔符，没有设置$IFS永久。这是因为输出行#2存在空格作为分隔符。
回答by gniourf_gniourf
Your first question is already addressed in F. Hauri's answer. Here's canonical way to join the elements of an array:
您的第一个问题已在F. Hauri 的回答中解决。这是连接数组元素的规范方法：
ids=( 1 2 3 4 )
IFS=\| eval 'lst="${ids[*]}"'
Some people will cry out loud that evalis evil, yet it's perfectly safe here, thanks to the single quotes. This only has advantages: there are no subshells, IFSis not globally modified, it will not trim trailing newlines, and it's very simple.
有些人会大声喊叫这eval是邪恶的，但由于单引号，这里非常安全。这只具有优点：没有子shell，IFS没有全局修改，不会修剪尾随的换行符，而且非常简单。
回答by codeforester
You can use printftoo, without any external commands or the need to manipulate IFS:
您也可以使用printf，无需任何外部命令或操作 IFS：
ids=(1 2 3 4) # create array
printf -v ids_d '|%s' "${ids[@]}" # yields "|1|2|3|4"
ids_d=${ids_d:1} # remove the leading '|'
回答by Léa Gris
An utility function to join arguments array into a delimited string:
将参数数组连接到分隔字符串的实用函数：
#!/usr/bin/env bash
# Join arguments with delimiter
# @Params
# : The delimiter string
# ${@:2}: The arguments to join
# @Output
# >&1: The arguments separated by the delimiter string
array::join() {
(($#)) || return 1 # At least delimiter required
local -- delim="" str IFS=
shift
str="${*/#/$delim}" # Expand arguments with prefixed delimiter (Empty IFS)
echo "${str:${#delim}}" # Echo without first delimiter
}
declare -a my_array=( 'Paris' 'Berlin' 'London' 'Brussel' 'Madrid' 'Oslo' )
array::join ', ' "${my_array[@]}"
array::join '*' {1..9} | bc # 1*2*3*4*5*6*7*8*9=362880 Factorial 9
declare -a null_array=()
array::join '== Ultimate separator of nothing ==' "${null_array[@]}"
Output:
输出：
Paris, Berlin, London, Brussel, Madrid, Oslo
362880
