如何将关联数组作为参数传递给 Bash 中的函数？
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How to pass an associative array as argument to a function in Bash?
提问by niksfirefly
How do you pass an associative array as an argument to a function? Is this possible in Bash?
如何将关联数组作为参数传递给函数？这在 Bash 中可能吗？
The code below is not working as expected:
下面的代码没有按预期工作：
function iterateArray
{
local ADATA="${@}" # associative array
for key in "${!ADATA[@]}"
do
echo "key - ${key}"
echo "value: ${ADATA[$key]}"
done
}
Passing associative arrays to a function like normal arrays does not work:
将关联数组传递给像普通数组这样的函数不起作用：
iterateArray "$A_DATA"
or
或者
iterateArray "$A_DATA[@]"
回答by Florian Feldhaus
I had exactly the same problem last week and thought about it for quite a while.
上周我遇到了完全相同的问题，并考虑了很长一段时间。
It seems, that associative arrays can't be serialized or copied. There's a good Bash FAQ entry to associative arrays which explains them in detail. The last section gave me the following idea which works for me:
看来，关联数组不能被序列化或复制。有一个很好的 Bash FAQ entry to associative arrays，其中详细解释了它们。最后一部分给了我以下对我有用的想法：
function print_array {
# eval string into a new associative array
eval "declare -A func_assoc_array="${1#*=}
# proof that array was successfully created
declare -p func_assoc_array
}
# declare an associative array
declare -A assoc_array=(["key1"]="value1" ["key2"]="value2")
# show associative array definition
declare -p assoc_array
# pass associative array in string form to function
print_array "$(declare -p assoc_array)"
回答by ling
Based on Florian Feldhaus's solution:
基于 Florian Feldhaus 的解决方案：
# Bash 4+ only
function printAssocArray # ( assocArrayName )
{
var=$(declare -p "")
eval "declare -A _arr="${var#*=}
for k in "${!_arr[@]}"; do
echo "$k: ${_arr[$k]}"
done
}
declare -A conf
conf[pou]=789
conf[mail]="ab\npo"
conf[doo]=456
printAssocArray "conf"
The output will be:
输出将是：
doo: 456
pou: 789
mail: ab\npo
回答by Todd Lehman
If you're using Bash 4.3 or newer, the cleanestway is to pass the associative array by name and then access it inside your function using a name reference with local -n. For example:
如果您使用的是 Bash 4.3 或更新版本，最简洁的方法是按名称传递关联数组，然后使用带有local -n. 例如：
function foo {
local -n data_ref=
echo ${data_ref[a]} ${data_ref[b]}
}
declare -A data
data[a]="Fred Flintstone"
data[b]="Barney Rubble"
foo data
You don't have to use the _refsuffix; that's just what I picked here. You can call the reference anything you want so long as it's different from the original variable name (otherwise youll get a "circular name reference" error).
您不必使用_ref后缀；这正是我在这里挑选的。只要它与原始变量名称不同，您就可以调用任何您想要的引用（否则您将收到“循环名称引用”错误）。
回答by Orwellophile
Update, to fully answer the question, here is an small section from my library:
更新，为了完全回答这个问题，这里是我图书馆的一小部分：
Iterating an associative array by reference
通过引用迭代关联数组
shopt -s expand_aliases
alias array.getbyref='e="$( declare -p )"; eval "declare -A E=${e#*=}"'
alias array.foreach='array.keys ; for key in "${KEYS[@]}"'
function array.print {
array.getbyref
array.foreach
do
echo "$key: ${E[$key]}"
done
}
function array.keys {
array.getbyref
KEYS=(${!E[@]})
}
# Example usage:
declare -A A=([one]=1 [two]=2 [three]=3)
array.print A
This we a devlopment of my earlier work, which I will leave below.
这是我早期工作的发展，我将在下面留下。
@ffeldhaus - nice response, I took it and ran with it:
@ffeldhaus - 很好的回应，我拿着它跑了：
t()
{
e="$( declare -p )"
eval "declare -A E=${e#*=}"
declare -p E
}
declare -A A='([a]="1" [b]="2" [c]="3" )'
echo -n original declaration:; declare -p A
echo -n running function tst:
t A
# Output:
# original declaration:declare -A A='([a]="1" [b]="2" [c]="3" )'
# running function tst:declare -A E='([a]="1" [b]="2" [c]="3" )'
回答by aks
You can only pass associative arrays by name.
您只能按名称传递关联数组。
It's better (more efficient) to pass regular arrays by name also.
最好（更有效）也按名称传递常规数组。
回答by Nickotine
yo:
哟：
#!/bin/bash
declare -A dict
dict=(
[ke]="va"
[ys]="lu"
[ye]="es"
)
fun() {
for i in $@; do
echo $i
done
}
fun ${dict[@]} # || ${dict[key]} || ${!dict[@] || ${dict[]}
eZ
eZ
回答by Samuel Powell
Here is a solution I came up with today using eval echo ...to do the indirection:
这是我今天提出的eval echo ...用于进行间接寻址的解决方案：
print_assoc_array() {
local arr_keys="${![@]}" # $ means we only substitute the
local arr_val="${[\"$k\"]}"
for k in $(eval echo $arr_keys); do #use eval echo to do the next substitution
printf "%s: %s\n" "$k" "$(eval echo $arr_val)"
done
}
declare -A my_arr
my_arr[abc]="123"
my_arr[def]="456"
print_assoc_array my_arr
Outputs on bash 4.3:
bash 4.3 上的输出：
def: 456
abc: 123
回答by l0b0
From the best Bash guideever:
来自有史以来最好的Bash 指南：
declare -A fullNames
fullNames=( ["lhunath"]="Maarten Billemont" ["greycat"]="Greg Wooledge" )
for user in "${!fullNames[@]}"
do
echo "User: $user, full name: ${fullNames[$user]}."
done
I think the issue in your case is that $@is notan associative array: "@: Expands to all the words of all the positional parameters. If double quoted, it expands to a list of all the positional parameters as individual words."
我觉得你的情况的问题是，$@是不是关联数组：“@：扩展到所有的位置参数的所有的话。如果双引号括起来，它扩展到所有的位置参数作为单个单词的列表。”
