如何在 Bash 中重命名关联数组？

Question

我需要遍历关联数组并将其内容导出到临时数组（并对值执行一些更新）。

然后应该丢弃第一个数组的剩余内容，我想将临时数组分配给原始数组变量。

须藤代码：

declare -A MAINARRAY
declare -A TEMPARRAY
... populate ${MAINARRAY[...]} ...

while something; do     #Drain some values from MAINARRAY to TEMPARRAY
    ${TEMPARRAY["$name"]}=((${MAINARRAY["$name"]} + $somevalue))
done
... other manipulations to TEMPARRAY ...

unset MAINARRAY        #discard left over values that had no update
declare -A MAINARRAY
MAINARRAY=${TEMPARRAY[@]}  #assign updated TEMPARRAY back to MAINARRAY (ERROR HERE)

Answer 1

这是用于任何类型的 bash 变量的小型复制函数
- 普通标量变量
- 索引数组
- 关联数组

### Function vcp    -VariableCoPy-  
# $1 Name of existing Source-Variable  
# $2 Name for the Copy-Target  
vcp() {
    local var=$(declare -p $1)
    var=${var/declare /declare -g }
    eval "${var/$1=/$2=}"
}

用法，例子：

# declarations
var="  345  89  "
ind_array=(Betty "  345  89  ")
declare -A asso_array=([one]=Harry [two]=Betty [some_signs]=" +*.<\$~,'/ ")  

# produce the copy
vcp var varcopy
vcp ind_array ind_array_copied
vcp asso_array asso_array_2   

# now you can check the equality between original and copy with commands like
# declare -p <name>

结果

--3    1: "${asso_array[@]}"   
(5)       asso_array[one]:        |Harry|   
(11)      asso_array[some_signs]: | +*.<$~,'/ |   
(5)       asso_array[two]:        |Betty|   
--3    4: "${asso_array_2[@]}"   
(5)       asso_array_2[one]:        |Harry|   
(11)      asso_array_2[some_signs]: | +*.<$~,'/ |   
(5)       asso_array_2[two]:        |Betty|   
--2    7: "${ind_array[@]}"   
(5)       ind_array[0]:   |Betty|   
(11)      ind_array[1]:   |  345  89  |   
--2    9: "${ind_array_copied[@]}"   
(5)       ind_array_copied[0]:   |Betty|   
(11)      ind_array_copied[1]:   |  345  89  |   
(11)  11: "$var":   |  345  89  |  
(11)  12: "$varcopy":   |  345  89  |  

Answer 2

这个怎么样（不创建真正的副本，只是指向源变量的链接）：

#!/bin/bash
declare -A my_array=(["key1"]="value1" ["key2"]="value2")
declare -n arr=my_array
arr['LOG_FILE']=/tmp/log.txt
echo ${arr['key1']}
echo ${arr['LOG_FILE']}

将打印：

value1
/tmp/log.txt

Answer 3

扩展Luca Borrione's cp_hash - 这对我不起作用，我放弃了尝试追踪 eval 扩展问题 - 我在 bash 4.2 之前和之后遇到了差异。在 4.2(something) 之后，这变得容易多了……但这不是向后兼容的。见1和2

所以我的变体在 4.1.2(1) 和 4.3.46(1) 上进行了测试：

#!/bin/bash
## bash4 due to associative arrays!

    function cp_hash() {
        ## REQUIRES you to declare -A $2 in advance.
        local original_hash_name="$1"
        local copy_hash_name="$2"
        #
        # sadly we have no way to identify if you have already declared it, so bull ahead.
        #
        ## store the definition of the old array
        local __copy__=$(declare -p $original_hash_name)
        ## rename the array inside the definition
        __copy__=${__copy__/${original_hash_name}=/__copy__=}

        ## for bash 4.2 > we could end here.
        ## declare -A creates local scope variables by default, so add -g
        ## this DOES NOT work prior to 4.2, even w/o -g and w/ a declare outside.
        #    __copy__=${__copy__/${original_hash_name}=/${copy_hash_name}=}
        #    eval ${__copy__/-A/-g -A}

        ## for bash4 where we can't do -g, then:
        ## local associative array based on the definition we stored and modified
        eval ${__copy__}
        ## loop through the local copy, and store it in the declared-outside copy.
        for i in "${!__copy__[@]}"
        do
            eval ${copy_hash_name}[$i]=${__copy__[$i]}
        done
    }

    declare -A hash
    hash[hello]=world
    hash[ab]=cd

    #not required for 4.2+ if you use -g, neither helps nor hinders
    declare -A copy

    cp_hash 'hash' 'copy'

    echo hash: ${hash[@]}
    echo copy: ${copy[@]}

    echo "copy result loop"
    for i in "${!copy[@]}"
    do
        echo "key  : $i | value: ${copy[$i]}"
    done

Answer 4

这个单行执行关联数组复制：MAINARRAY=TEMPARRAY

eval $(typeset -A -p TEMPARRAY|sed 's/ TEMPARRAY=/ MAINARRAY=/')

Answer 5

根据glenn jackman 和ffeldhaus 的建议，您可以构建一个可能会变得方便的函数：

function cp_hash
{
    local original_hash_name="$1"
    local copy_hash_name="$2"

    local __copy__=$(declare -p $original_hash_name);
    eval declare -A __copy__="${__copy__:$(expr index "${__copy__}" =)}";

    for i in "${!__copy__[@]}"
    do
        eval ${copy_hash_name}[$i]=${__copy__[$i]}
    done
}

用法：

declare -A copy_hash_name
cp_hash 'original_hash_name' 'copy_hash_name'

示例：

declare -A hash
hash[hello]=world
hash[ab]=cd

declare -A copy
cp_hash 'hash' 'copy'

for i in "${!copy[@]}"
do
    echo "key  : $i | value: ${copy[$i]}"
done

会输出

key  : ab | value: cd
key  : hello | value: world

Answer 6

在 bash 中无法直接复制关联数组。正如已经指出的那样，最好的解决方案可能是遍历数组并逐步复制它。

还有另一个解决方案I used to pass variables to functions。您可以使用相同的技术来复制关联数组：

# declare associative array
declare -A assoc_array=(["key1"]="value1" ["key2"]="value2")
# convert associative array to string
assoc_array_string=$(declare -p assoc_array)
# create new associative array from string
eval "declare -A new_assoc_array="${assoc_array_string#*=}
# show array definition
declare -p new_assoc_array

Answer 7

对于关联数组，我认为除了迭代之外没有其他方法

for key in "${!TEMPARRAY[@]}"  # make sure you include the quotes there
do
  MAINARRAY["$key"]="${TEMPARRAY["$key"]}"
  # or: MAINARRAY+=( ["$key"]="${TEMPARRAY["$key"]}" )
done

Answer 8

MAINARRAY=( "${TEMPARRAY[@]}" )