php - PHP array_intersect_key但保留array2的值

我有2个数组：

$array1 = ["b" => "no", "c" => "no", "d" => ["y" => "no"]];
$array2 = ["a" => "yes", "b" => "yes", "c" => "yes", "d" => ["x" => "yes", "y" => "yes", "z" => "yes"]];

我想返回$array2的成员，这些成员的键也出现在$array1中。

所需的输出：

$array3 = ["b" => "yes", "c" => "yes", "d" => ["y" => "yes"]];

基本上，我需要与array_intersect_key相同的功能，但我想保留$array2而不是$array1的值。

是否存在这样的功能？还需要递归到子数组中。谢谢！

*编辑*
正如安德里（Andri）在his comment中所建议的那样，我试图翻转array_intersect_key($array2, $array1)周围的参数，但在这种情况下，我真正得到的是：

$array3 = ["b" => "yes", "c" => "yes", "d" => ["x" => "yes", "y" => "yes", "z" => "yes"]];

即“ d”的所有子元素，只是因为在$array1中提到了“ d”。

最佳答案

据我所知，没有内置函数，使用递归处理所有层的方法应该很容易...

$array1 = ["b" => "no", "c" => "no", "d" => ["y" => "no"]];
$array2 = ["a" => "yes", "b" => "yes", "c" => "yes", "d" => ["x" => "yes", "y" => "yes", "z" => "yes"]];

function intersect_key_2 ( array $a1, array $a2 )   {
    foreach ( $a1 as $key1 => $value1)  {
        // If there is a matching element
        if ( isset($a2[$key1]) )    {
            // Update - if an array, use recursive call else just replace
            $a1[$key1] = ( is_array($value1) ) ? intersect_key_2 ( $value1, $a2[$key1])
                    : $a2[$key1];
        }
    }

    return $a1;
}

print_r(intersect_key_2($array1, $array2));

给

Array
(
    [b] => yes
    [c] => yes
    [d] => Array
        (
            [y] => yes
        )

)

关于php - PHP array_intersect_key但保留array2的值，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/59107175/