PHP获取给定目录的所有子目录

Question

如何在没有文件.(当前目录)或..(父目录)的情况下获取给定目录的所有子目录,然后使用函数中的每个目录？

Answer 1

你可以使用glob()和GLOB_ONLYDIR选项

要么

$dirs = array_filter(glob('*'), 'is_dir');
print_r( $dirs);

Answer 2

以下是如何使用GLOB仅检索目录:

$directories = glob($somePath . '/*' , GLOB_ONLYDIR);

Answer 3

Spl DirectoryIterator类提供了一个用于查看文件系统目录内容的简单界面.

$dir = new DirectoryIterator($path);
foreach ($dir as $fileinfo) {
    if ($fileinfo->isDir() && !$fileinfo->isDot()) {
        echo $fileinfo->getFilename().'<br>';
    }
}

Answer 4

与上一个问题几乎相同:

$iterator = new RecursiveIteratorIterator(
                new RecursiveDirectoryIterator($yourStartingPath), 
            RecursiveIteratorIterator::SELF_FIRST);

foreach($iterator as $file) {
    if($file->isDir()) {
        echo strtoupper($file->getRealpath()), PHP_EOL;
    }
}

替换strtoupper为您想要的功能.

Answer 5

在数组中：

function expandDirectoriesMatrix($base_dir, $level = 0) {
    $directories = array();
    foreach(scandir($base_dir) as $file) {
        if($file == '.' || $file == '..') continue;
        $dir = $base_dir.DIRECTORY_SEPARATOR.$file;
        if(is_dir($dir)) {
            $directories[]= array(
                    'level' => $level
                    'name' => $file,
                    'path' => $dir,
                    'children' => expandDirectoriesMatrix($dir, $level +1)
            );
        }
    }
    return $directories;
}

//使用权：

$dir = '/var/www/';
$directories = expandDirectoriesMatrix($dir);

echo $directories[0]['level']                // 0
echo $directories[0]['name']                 // pathA
echo $directories[0]['path']                 // /var/www/pathA
echo $directories[0]['children'][0]['name']  // subPathA1
echo $directories[0]['children'][0]['level'] // 1
echo $directories[0]['children'][1]['name']  // subPathA2
echo $directories[0]['children'][1]['level'] // 1

显示所有示例：

function showDirectories($list, $parent = array())
{
    foreach ($list as $directory){
        $parent_name = count($parent) ? " parent: ({$parent['name']}" : '';
        $prefix = str_repeat('-', $directory['level']);
        echo "$prefix {$directory['name']} $parent_name <br/>";  // <-----------
        if(count($directory['children'])){
            // list the children directories
            showDirectories($directory['children'], $directory);
        }
    }
}

showDirectories($directories);

// pathA
// - subPathA1 (parent: pathA)
// -- subsubPathA11 (parent: subPathA1)
// - subPathA2 
// pathB
// pathC

Answer 6

试试这段代码:

<?php
$path = '/var/www/html/project/somefolder';

$dirs = array();

// directory handle
$dir = dir($path);

while (false !== ($entry = $dir->read())) {
    if ($entry != '.' && $entry != '..') {
       if (is_dir($path . '/' .$entry)) {
            $dirs[] = $entry; 
       }
    }
}

echo "<pre>"; print_r($dirs); exit;