bash - 如何使用命令输出填充 bash 关联数组？

Question

我正在尝试用命令的输出填充关联数组。我可以在没有命令的情况下做到这一点:

$ declare -A x=( [first]=foo [second]=bar )
$ echo "${x[first]}, ${x[second]}"
foo, bar

我可以用命令输出填充一个非关联数组:

$ declare y=( $(echo 'foo bar') )
$ echo "${y[0]}, ${y[1]}"
foo, bar

但是当我尝试在上述两个基础上创建一个将从命令填充关联数组的语句时，我收到以下错误消息:

$ declare -A z=( $(echo '[first]=foo [second]=bar') )
-bash: z: $(echo '[first]=foo [second]=bar'): must use subscript when assigning associative array

为什么我会收到该错误消息？使用命令输出填充关联数组的正确语法是什么？我试图避免对 the usual reasons 使用 eval , 不想使用临时文件，当然 echo 只是用作产生问题效果的命令的示例，真正的命令会更复杂。

因此，根据下面的几个答案，似乎只是我的引用有问题:

$ declare -A z="( $(echo '[first]=foo [second]=bar') )"
$ echo "${z[first]}, ${z[second]}"
foo, bar

索引和值中有空格:

$ declare -A z="( $(echo '[first field]="foo with space" [second]="space bar"') )"
$ echo "${z[first field]}, ${z[second]}"
foo with space, space bar

---

编辑以回应评论中关于为什么需要引号的问题(How do I populate a bash associative array with command output?)-我不完全知道，但也许其他人可以解释使用此脚本的结果作为引用(不期望指定的索引要在索引数组中使用，它们只是作为数组值填充的字符串的一部分):

$ cat tst.sh
#!/bin/env bash

set -x

printf 'Indexed, no quotes\n'
declare -a w=( $(echo '[first]=foo [second]=bar') )
declare -p w

printf '\n---\n'

printf 'Indexed, with quotes\n'
declare -a x="( $(echo '[first]=foo [second]=bar') )"
declare -p x

printf '\n---\n'

printf 'Associative, no quotes\n'
declare -A y="( $(echo '[first]=foo [second]=bar') )"
declare -p y

printf '\n---\n'

printf 'Associative, with quotes\n'
declare -A z=( $(echo '[first]=foo [second]=bar') )
declare -p z

.

$ ./tst.sh
+ printf 'Indexed, no quotes\n'
Indexed, no quotes
+ w=($(echo '[first]=foo [second]=bar'))
++ echo '[first]=foo [second]=bar'
+ declare -a w
+ declare -p w
declare -a w=([0]="[first]=foo" [1]="[second]=bar")
+ printf '\n---\n'

---
+ printf 'Indexed, with quotes\n'
Indexed, with quotes
++ echo '[first]=foo [second]=bar'
+ declare -a 'x=( [first]=foo [second]=bar )'
+ declare -p x
declare -a x=([0]="bar")
+ printf '\n---\n'

---
+ printf 'Associative, no quotes\n'
Associative, no quotes
++ echo '[first]=foo [second]=bar'
+ declare -A 'y=( [first]=foo [second]=bar )'
+ declare -p y
declare -A y=([second]="bar" [first]="foo" )
+ printf '\n---\n'

---
+ printf 'Associative, with quotes\n'
Associative, with quotes
+ z=($(echo '[first]=foo [second]=bar'))
./tst.sh: line 24: z: $(echo '[first]=foo [second]=bar'): must use subscript when assigning associative array
+ declare -A z
+ declare -p z
declare -A z=()

Answer 1

这是一种传统的 while 循环方法，用于从命令的输出中填充关联数组:

while IFS= read -r; do
   declare -A z+="( $REPLY )"
done < <(printf '[first]=foo [second]=bar\n[third]=baz\n')

# check output
$> echo "${z[first]}, ${z[second]}, ${z[third]}"
foo, bar, baz

# or declare -p
$> declare -p z
declare -A z='([third]="baz" [second]="bar" [first]="foo" )'

编辑:您最初的尝试也适用于正确的引号:

$> unset z

$> declare -A z="( $(echo '[first]=foo [second]=bar') )"

$> declare -p z
declare -A z='([second]="bar" [first]="foo" )'