Question

如何使它成为array_merge（）用两个数组覆盖具有不同值但相同键索引的两个键？

例如，合并：

[0] => 'whatever'

与

[0] => 'whatever', [1] => 'a', [2] => 'b'

应该产生

[0] => 'whatever', [1] => 'a', [2] => 'b'

基本上我希望array_merge与数组有字符串键时的行为方式相同...

Answer 1

使用+运算符。

将array_merge与+运算符进行比较：

<?php

$a1 = array(0=>"whatever",);
$a2 = array(0=>"whatever",1=>"a",2=>"b");

print_r(array_merge($a1,$a2));
print_r($a1+$a2);
?>

输出：

Array
(
    [0] => whatever
    [1] => whatever
    [2] => a
    [3] => b
)
Array
(
    [0] => whatever
    [1] => a
    [2] => b
)

如果关联数组的数字键无序，则+运算符仍然有效：

<?php

$a1 = array(0=>"whatever",);
$a2 = array(1=>"a",0=>"whatever",2=>"b");

print_r(array_merge($a1,$a2));
print_r($a1+$a2);
?>

输出：

Array
(
    [0] => whatever
    [1] => a
    [2] => whatever
    [3] => b
)
Array
(
    [0] => whatever
    [1] => a
    [2] => b
)

注意此案例中的array_merge会创建一个新密钥。不可取......

Answer 2

手动编写非常简单：

function array_merge_custom($first, $second) {
    $result = array();
    foreach($first as $key => $value) {
        $result[$key] = $value;
    }
    foreach($second as $key => $value) {
        $result[$key] = $value;
    }

    return $result;
}

更新：这与union运算符（return $first + $second;）的行为不同，因为在这种情况下， second 数组在两个元素具有相同键时都会获胜。

但是，如果切换参数的位置并将要“赢”的数组放在第一个操作数的冲突情况下，则可以获得相同的行为。因此，上述函数的行为与return $second + $first;完全相同。

Answer 3

array_replace就是这样做的。请参阅：http://php.net/manual/de/function.array-replace.php

Answer 4

在我的项目中，我使用自己的功能

function array_merge_custom(){
    $array = [];
    $arguments  = func_num_args();
    foreach($arguments as $args)
        foreach($args as $key => $value)
            $array[$key] = $value;
    return $array;
}

用法

$a = array_merge_custom($b, $c, $d, ... .. )

Answer 5

您应该使用$a2+$a1与array_merge($a1,$a2);

获得相同的结果

$a1 = array(
    'k1' => 1,
    'k2' => 2,
    'k3' => 3,
);

$a2 = array(
    'k1' => 11,
    'k2' => 22,
    'k4' => 44,
);

<强>代码：

print_r(array_merge($a1,$a2));

输出

Array ( 
    [k1] => 11 
    [k2] => 22 
    [k3] => 3 
    [k4] => 44 
)

<强>代码：

print_r($a1+$a2);

输出

Array ( 
    [k1] => 1 
    [k2] => 2 
    [k3] => 3 
    [k4] => 44 
)

<强>代码：

print_r($a2+$a1);

输出

Array ( 
    [k1] => 11 
    [k2] => 22 
    [k4] => 44 
    [k3] => 3 
)

Answer 6

您可以使用array_merge()，然后使用array_unique()。

Answer 7

the solution could be this:
function array_merge_custom($array1, $array2) {
    $mergeArray = [];
    $array1Keys = array_keys($array1);
    $array2Keys = array_keys($array2);
    $keys = array_merge($array1Keys, $array2Keys);

    foreach ($keys as $key) {
        $mergeArray[$key] = array_merge_recursive(isset($array1[$key]) ? $array1[$key] : [], isset($array2[$key]) ? $array2[$key] : []);
    }

    return $mergeArray;
}

$array1 = [
    '66_' => [
        'k1' => 1,
        'k2' => 1,
    ],
    '67_' => [
        'k1' => 1,
        'k2' => 1,
    ],
    '68_' => [
        'k1' => 1,
        'k2' => 1,
    ],
    68 => [
        'k1' => 1,
        'k2' => 1,
    ]
];
$array2 = [
    '66_' => [
        'a1' => 1,
        'a2' => 1,
    ],
    '68_' => [
        'b1' => 1,
        'b2' => 1,
    ],
    68 => [
        'b1' => 1,
        'b2' => 1,
    ]
];
echo '<pre>';
print_r(array_merge_custom($array1, $array2));

Answer 8

$arrA = [10, 11, 12];
$arrB = [12, 13];

$arrCommon = array_keys(array_flip($arrA) + array_flip($arrB));

print_r($arrCommon);
Array
(
    [0] => 10
    [1] => 11
    [2] => 12
    [3] => 13
)

与错误使用＆＃34; +＆＃34;

进行比较

$arrCommon = $arrA + $arrB;

print_r($arrCommon);
Array
(
    [0] => 10
    [1] => 11
    [2] => 12
)

带有数字键的PHP array_merge

8 个答案: