如何将关联数组作为参数传递给Bash中的函数?

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是行不通的:

iterateArray "$A_DATA"

or

或

iterateArray "$A_DATA[@]"

7 个解决方案

#1

I had exactly the same problem last week and thought about it for quite a while.

上周我遇到了同样的问题，想了很久。

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:

似乎关联数组不能序列化或复制。有一个很好的关于关联数组的Bash FAQ条目，它详细地解释了它们。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)" 

#2

Based on Florian Feldhaus's solution:

根据Florian Feldhaus的解决方案:

# Bash 4+ only
function printAssocArray # ( assocArrayName ) 
{
    var=$(declare -p "$1")
    eval "declare -A _arr="${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:

的输出将会是:

doo: 456
pou: 789
mail: ab\npo

#3

Update, to fully answer the question, here is an small section from my library:

更新，充分回答这个问题，以下是我的图书馆的一小部分:

Iterating an associative array by reference

shopt -s expand_aliases
alias array.getbyref='e="$( declare -p ${1} )"; eval "declare -A E=${e#*=}"'
alias array.foreach='array.keys ${1}; for key in "${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.

这是对我早期作品的总结，我将在下面留下。

@ffeldhaus - nice response, I took it and ran with it:

@ffeldhaus -回复不错，我拿着它跑了:

t() 
{
    e="$( declare -p $1 )"
    eval "declare -A E=${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" )'

#4

You can only pass associative arrays by name.

只能通过名称传递关联数组。

It's better (more efficient) to pass regular arrays by name also.

通过名称传递常规数组更好(更有效)。

#5

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[$1]} 

eZ

易之

#6

Here is a solution I came up with today using eval echo ... to do the indirection:

下面是我今天使用eval echo提出的解决方案。间接的:

print_assoc_array() {
    local arr_keys="\${!$1[@]}" # \$ means we only substitute the $1
    local arr_val="\${$1[\"\$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:

在bash 4.3输出:

def: 456
abc: 123

#7

From the best Bash guide ever:

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 not an 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."

我认为您的问题是$@不是一个关联数组:“@:扩展到所有位置参数的所有单词。如果双引号，它将扩展为所有位置参数作为单个单词的列表。