将 Bash 数组转换为分隔字符串 [英] Converting a Bash array into a delimited string

问题描述

我想知道以下内容；

1. 为什么给定的非工作示例不起作用.
2. 是否还有其他比工作示例中给出的方法更简洁的方法.

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

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

在上下文中，要在 sed 中使用的分隔字符串进一步解析的命令.

In context, the delimited string to be used in a sed command for further parsing.

推荐答案

# REVISION: 2017-03-14
# Use of read and other bash specific features (bashisms)

因为括号用于分隔数组，而不是字符串:

Because parentheses are used to delimit an array, not a string:

ids="1 2 3 4";echo ${ids// /|}
1|2|3|4

一些示例:用两个字符串填充 $ids:a b 和 c d

Some samples: Populating $ids with two strings: a b and c d

ids=("a b" "c d")

echo ${ids[*]// /|}
a|b c|d

IFS='|';echo "${ids[*]}";IFS=$' \t\n'
a b|c d

...最后:

IFS='|';echo "${ids[*]// /|}";IFS=$' \t\n'
a|b|c|d

组装数组的地方，由 $IFS 的第一个字符分隔，但在数组的每个元素中用 | 替换空格.

Where array is assembled, separated by 1st char of $IFS, but with space replaced by | in each element of array.

当你这样做时:

id="${ids[@]}"

您将 array ids 通过空格合并后的字符串构建转移到 string 类型的新变量.

you transfer the string build from the merging of the array ids by a space to a new variable of type string.

注意:当"${ids[@]}"给出一个空格分隔的字符串时，"${ids[*]}"(使用星号 * 而不是 at 符号 @)将呈现由 的 第一个字符分隔的字符串>$IFS.

Note: when "${ids[@]}" give a space-separated string, "${ids[*]}" (with a star * instead of the at sign @) will render a string separated by the first character of $IFS.

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>''.

玩$IFS:

set | grep ^IFS=
IFS=$' \t\n'

declare -p IFS
declare -- IFS=" 
"
printf "%q\n" "$IFS"
$' \t\n'

字面意思是一个空格、一个制表和(意思是或)一个换行.所以，虽然第一个字符是一个空格.* 的使用与 @ 的作用相同.

Literally a space, a tabulation and (meaning or) a line-feed. So, while the first character is a space. the use of * will do the same as @.

但是:

{

    # 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

注意:行 IFS=: read -a array <;<(...) 将使用 : 作为分隔符，而不永久设置 $IFS.这是因为输出行 #2 将空格作为分隔符.

Note: The line IFS=: read -a array < <(...) will use : as separator, without setting $IFS permanently. This is because output line #2 present spaces as separators.

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