array(
    'data1' => <some data>,'next' => array(
        'data2' => <some data>,'next' => array(
            'data3' => <some data>,'next' => array(
                'data4' => <some data>,'next' => array(
                    'data5' => <some data>,'next' => ..... etc to data n
                );
            );
        );
    );
);

你不是在寻找阵列的反面,但是你正在寻找相反的东西.

首先要更好地理解这种逆转可能会对你有所帮助.

您需要每个元素的父元素.如果您遍历下一个,则父级始终是前一个.因此,如果您将前一个元素添加为父元素,那么数组的最后一个元素就是您要查找的元素.

所以听起来很直接.更难的部分是用所谓的变量别名/引用来表达它.

让我们添加所有父项,并在遍历数组时,在引用它之后删除“下一个”条目：

/* traverse the path on 'next' and keep prevIoUs to set 'parent' of current */
$walk = &$array; // start at root node
while ($walk) {

    if (isset($PRevIoUs)) {
        $walk['parent'] = &$prevIoUs;
    }

    $prevIoUs = &$walk;

    $hasNext = array_key_exists('next',$walk);
    if ($hasNext) {
        $walk = &$walk['next'];
        unset($prevIoUs['next']);
    } else {
        break;
    }
}
unset($prevIoUs);

写入最后一个元素然后将包含您正在寻找的数组.最后一个元素是$步行：

print_r($walk);

哪个给你(Demo)：

Array
(
    [data5] => <some data5>
    [parent] => Array
        (
            [data4] => <some data4>
            [parent] => Array
                (
                    [data3] => <some data3>
                    [parent] => Array
                        (
                            [data2] => <some data2>
                            [parent] => Array
                                (
                                    [data1] => <some data1>
                                )
    ...
)

希望这是有帮助的,可以理解的.