Php 合并并覆盖两个或多个多维数组

如何使用array\u merge\u recursive合并和覆盖具有相同键和值的多维数组？假设我有两个数组，如下所示：

$arr1 = array(       
   // OVERWRITE
   array('prop_id' => 1, 'prop_value' => 'batman'),
   array('prop_id' => 2, 'prop_value' => 'ironman'),
   // NOT OVERWRITE
   array('prop_id' => 5, 'prop_value' => 'wonderwoman'),
);

$arr2 = array(
   array('prop_id' => 1, 'prop_value' => 'robin'),
   array('prop_id' => 2, 'prop_value' => 'superman'),
   array('prop_id' => 4, 'prop_value' => 'catwoman'),
);

$result = array_replace_recursive(
    array(
       1 => array('prop_id' => 1, 'prop_value' => 'batman'),
       2 => array('prop_id' => 2, 'prop_value' => 'ironman'),
       5 => array('prop_id' => 5, 'prop_value' => 'wonderwoman'),
    ),
    array(
       1 => array('prop_id' => 1, 'prop_value' => 'robin'),
       2 => array('prop_id' => 2, 'prop_value' => 'superman'),
       4 => array('prop_id' => 4, 'prop_value' => 'catwoman'),
    ),
);

我想合并并用新值覆盖它（规则是比较键，与它没有覆盖的值相同），预期结果是

 $result = array_merge_overwrite($arr1, $arr2, array('prop_id') /* Comparison Key */);
 $result = array(       
   array('prop_id' => 1 /* Comparison Key */, 'prop_value' => 'robin' /* Comparison value */),
   array('prop_id' => 2, 'prop_value' => 'superman'),
   array('prop_id' => 4, 'prop_value' => 'catwoman'),
   array('prop_id' => 5, 'prop_value' => 'wonderwoman'),
);

使用array\u merge\u recursive，它不会被覆盖，我尝试使用array\u replace\u recursive，如下所示：

$arr1 = array(       
   // OVERWRITE
   array('prop_id' => 1, 'prop_value' => 'batman'),
   array('prop_id' => 2, 'prop_value' => 'ironman'),
   // NOT OVERWRITE
   array('prop_id' => 5, 'prop_value' => 'wonderwoman'),
);

$arr2 = array(
   array('prop_id' => 1, 'prop_value' => 'robin'),
   array('prop_id' => 2, 'prop_value' => 'superman'),
   array('prop_id' => 4, 'prop_value' => 'catwoman'),
);

$result = array_replace_recursive(
    array(
       1 => array('prop_id' => 1, 'prop_value' => 'batman'),
       2 => array('prop_id' => 2, 'prop_value' => 'ironman'),
       5 => array('prop_id' => 5, 'prop_value' => 'wonderwoman'),
    ),
    array(
       1 => array('prop_id' => 1, 'prop_value' => 'robin'),
       2 => array('prop_id' => 2, 'prop_value' => 'superman'),
       4 => array('prop_id' => 4, 'prop_value' => 'catwoman'),
    ),
);

---

这是可行的，但我的代码看起来又脏又脏。如我所见，任何比我的更好的解决方案都是，您希望保留第二个数组元素，以防第一个数组和第二个数组的键匹配

PHP手册中说： 如果要将第二个数组中的数组元素追加到第一个数组中，同时不覆盖第一个数组中的元素，也不重新编制索引，请使用+数组联合运算符：

<?php
$array1 = array(0 => 'zero_a', 2 => 'two_a', 3 => 'three_a');
$array2 = array(1 => 'one_b', 3 => 'three_b', 4 => 'four_b');
$result = $array1 + $array2;
var_dump($result);
?> 

---

有关更多详细信息，请查看，正如我所看到的，您希望保留第二个数组元素，以防第一个数组和第二个数组的键匹配

PHP手册中说： 如果要将第二个数组中的数组元素追加到第一个数组中，同时不覆盖第一个数组中的元素，也不重新编制索引，请使用+数组联合运算符：

<?php
$array1 = array(0 => 'zero_a', 2 => 'two_a', 3 => 'three_a');
$array2 = array(1 => 'one_b', 3 => 'three_b', 4 => 'four_b');
$result = $array1 + $array2;
var_dump($result);
?> 

---

有关更多详细信息，请查看以下功能，该功能将按照您的描述工作：

function array_merge_overwrite($arr1, $arr2, $uniques=array('prop_id'), $delimiter='-')
{
    $result = array();
    $uk = array();
    foreach($arr1 as $a1)
    {   
        $uk = array();
        foreach($uniques as $u) $uk[] = $a1[$u];
        $result[implode($delimiter, $uk)] = $a1;
    }   

    foreach($arr2 as $a2)
    {   
        $uk = array();
        foreach($uniques as $u) $uk[] = $a2[$u];
        $result[implode($delimiter, $uk)] = $a2;
    }   
    return $result;
}

如果按照问题中的定义传递了

$arr1

和

$arr2

，则上述函数将返回一个数组：

Array
(
    [1] => Array
        (
            [prop_id] => 1
            [prop_value] => robin
        )

    [2] => Array
        (
            [prop_id] => 2
            [prop_value] => superman
        )

    [5] => Array
        (
            [prop_id] => 5
            [prop_value] => wonderwoman
        )

    [4] => Array
        (
            [prop_id] => 4
            [prop_value] => catwoman
        )

)

当然，如果始终且仅使用

prop\u id

作为唯一元素，那么函数可能会简单一些：

function array_merge_overwrite($arr1, $arr2)
{
    $tmp = array();
    foreach($arr1 as $a1) $tmp[$a1['prop_id']] = $a1['prop_value'];
    foreach($arr2 as $a2) $tmp[$a2['prop_id']] = $a2['prop_value'];
    $result = array();
    foreach($tmp as $k=>$v) $result[] = array('prop_id'=>$k, 'prop_value'=>$v);
    return $result;
}

---

在后面的函数中，返回数组的唯一区别在于元素数组的键将是标准数值，而不是与

属性id

s

匹配。下面是一个按您所述工作的函数：

function array_merge_overwrite($arr1, $arr2, $uniques=array('prop_id'), $delimiter='-')
{
    $result = array();
    $uk = array();
    foreach($arr1 as $a1)
    {   
        $uk = array();
        foreach($uniques as $u) $uk[] = $a1[$u];
        $result[implode($delimiter, $uk)] = $a1;
    }   

    foreach($arr2 as $a2)
    {   
        $uk = array();
        foreach($uniques as $u) $uk[] = $a2[$u];
        $result[implode($delimiter, $uk)] = $a2;
    }   
    return $result;
}

如果按照问题中的定义传递了

$arr1

和

$arr2

，则上述函数将返回一个数组：

Array
(
    [1] => Array
        (
            [prop_id] => 1
            [prop_value] => robin
        )

    [2] => Array
        (
            [prop_id] => 2
            [prop_value] => superman
        )

    [5] => Array
        (
            [prop_id] => 5
            [prop_value] => wonderwoman
        )

    [4] => Array
        (
            [prop_id] => 4
            [prop_value] => catwoman
        )

)

当然，如果始终且仅使用

prop\u id

作为唯一元素，那么函数可能会简单一些：

function array_merge_overwrite($arr1, $arr2)
{
    $tmp = array();
    foreach($arr1 as $a1) $tmp[$a1['prop_id']] = $a1['prop_value'];
    foreach($arr2 as $a2) $tmp[$a2['prop_id']] = $a2['prop_value'];
    $result = array();
    foreach($tmp as $k=>$v) $result[] = array('prop_id'=>$k, 'prop_value'=>$v);
    return $result;
}

在后面的函数中，返回数组的唯一区别在于元素数组的键将是标准数值，而不是匹配

prop\u id

s