Php 如何根据允许用户查看的文件构造此文件树？

我有一组文件，如下所示：

Array
(
    [0] => Array
        (
            [type] => folder
            [path] => RootFolder
        )

    [1] => Array
        (
            [type] => file
            [path] => RootFolder\error.log
        )

    [2] => Array
        (
            [type] => folder
            [path] => RootFolder\test
        )

    [3] => Array
        (
            [type] => file
            [path] => RootFolder\test\asd.txt
        )

    [4] => Array
        (
            [type] => folder
            [path] => RootFolder\test\sd
        )

    [5] => Array
        (
            [type] => file
            [path] => RootFolder\test\sd\testing.txt
        )
)

RootFolder
    - error.log
    - test
        - asd.txt
        - sd
            - testing.txt

Array
(
    [0] => Array
        (
            [filePath] => RootFolder\test\sd
        )

    [1] => Array
        (
            [filePath] => RootFolder\error.log
        )

)

我解析这个数组，并根据文件的深度（“/”计数）创建一个树状视图。看起来是这样的：

Array
(
    [0] => Array
        (
            [type] => folder
            [path] => RootFolder
        )

    [1] => Array
        (
            [type] => file
            [path] => RootFolder\error.log
        )

    [2] => Array
        (
            [type] => folder
            [path] => RootFolder\test
        )

    [3] => Array
        (
            [type] => file
            [path] => RootFolder\test\asd.txt
        )

    [4] => Array
        (
            [type] => folder
            [path] => RootFolder\test\sd
        )

    [5] => Array
        (
            [type] => file
            [path] => RootFolder\test\sd\testing.txt
        )
)

RootFolder
    - error.log
    - test
        - asd.txt
        - sd
            - testing.txt

Array
(
    [0] => Array
        (
            [filePath] => RootFolder\test\sd
        )

    [1] => Array
        (
            [filePath] => RootFolder\error.log
        )

)

我现在拥有的是一组允许用户查看的文件路径。在构建上面的树时，我需要考虑这个数组。该数组如下所示：

Array
(
    [0] => Array
        (
            [type] => folder
            [path] => RootFolder
        )

    [1] => Array
        (
            [type] => file
            [path] => RootFolder\error.log
        )

    [2] => Array
        (
            [type] => folder
            [path] => RootFolder\test
        )

    [3] => Array
        (
            [type] => file
            [path] => RootFolder\test\asd.txt
        )

    [4] => Array
        (
            [type] => folder
            [path] => RootFolder\test\sd
        )

    [5] => Array
        (
            [type] => file
            [path] => RootFolder\test\sd\testing.txt
        )
)

RootFolder
    - error.log
    - test
        - asd.txt
        - sd
            - testing.txt

Array
(
    [0] => Array
        (
            [filePath] => RootFolder\test\sd
        )

    [1] => Array
        (
            [filePath] => RootFolder\error.log
        )

)

如果在数组（$path，$allowed）中执行

将很容易

，但这不会给出树。只是一个文件列表

另一个让我费解的部分是这个要求：如果用户有权查看文件夹

test

，那么他们就有权访问该文件夹的所有子文件夹

我的想法是简单地解析文件路径。例如，我要确认

RootFolder\test\sd

是一个目录，然后根据“/”计数创建一个树。就像我之前做的那样。然后，由于这是一个目录，我会取出这个目录中的所有文件，并将它们显示给用户。但是，我无法将其转换为工作代码

有什么想法吗

$tree = array();
// keep one:
$permNeeded = '?'; //something you're searching for exactly
$permNeeded = array('?', '?'); // multiple allowed perms
// be carefull with octal data checking permisions!

function checkPerms($permFileHas){
    // keep one:
    return $permFileHas==$permNeeded;
    return in_array($permFileHas, $permNeeded);
}

function parseDir($dir){
    $contents = scandir($dir);
    foreach($contents as $file){
        if(in_array($file, array('.', '..')){
            continue; // skip . and ..
        }
        if(is_dir($file)){
            parseDir($file);
            continue;
        }
        if(checkPerms(fileperms($file)){
            $tree[] = $dir.DIRECTORY_SEPARATOR.$file;
        }
    }
}

parseDir('/the/dir/user/have/perms');

这应该可以做到：）