如何获取没有文件的给定目录的所有子目录.（当前目录）或..（父目录），然后在函数中使用每个目录？
问问题
234346 次
16 回答
232
选项1：
您可以使用glob()该GLOB_ONLYDIR选项。
选项 2：
另一种选择是用于array_filter过滤目录列表。但是，请注意，下面的代码将跳过名称中带有句点的有效目录，例如.config.
$dirs = array_filter(glob('*'), 'is_dir');
print_r($dirs);
于 2010-03-26T14:58:34.287 回答
174
以下是使用 GLOB 仅检索目录的方法：
$directories = glob($somePath . '/*' , GLOB_ONLYDIR);
于 2010-06-09T13:44:29.540 回答
59
Spl DirectoryIterator类提供了一个简单的界面来查看文件系统目录的内容。
$dir = new DirectoryIterator($path);
foreach ($dir as $fileinfo) {
if ($fileinfo->isDir() && !$fileinfo->isDot()) {
echo $fileinfo->getFilename().'<br>';
}
}
于 2013-10-04T10:14:38.257 回答
29
几乎与您之前的问题相同：
$iterator = new RecursiveIteratorIterator(
new RecursiveDirectoryIterator($yourStartingPath),
RecursiveIteratorIterator::SELF_FIRST);
foreach($iterator as $file) {
if($file->isDir()) {
echo strtoupper($file->getRealpath()), PHP_EOL;
}
}
替换strtoupper为您想要的功能。
于 2010-03-26T15:11:12.670 回答
7
在数组中：
function expandDirectoriesMatrix($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
于 2017-03-09T10:24:42.437 回答
6
试试这个代码：
<?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;
于 2015-11-06T07:11:16.323 回答
3
你可以试试这个功能（需要PHP 7）
function getDirectories(string $path) : array
{
$directories = [];
$items = scandir($path);
foreach ($items as $item) {
if($item == '..' || $item == '.')
continue;
if(is_dir($path.'/'.$item))
$directories[] = $item;
}
return $directories;
}
于 2016-09-12T09:44:12.633 回答
3
非递归仅列出目录
唯一直接问这个的问题被错误地关闭了，所以我必须把它放在这里。
它还提供了过滤目录的能力。
/**
* Copyright © 2020 Theodore R. Smith <https://www.phpexperts.pro/>
* License: MIT
*
* @see https://stackoverflow.com/a/61168906/430062
*
* @param string $path
* @param bool $recursive Default: false
* @param array $filtered Default: [., ..]
* @return array
*/
function getDirs($path, $recursive = false, array $filtered = [])
{
if (!is_dir($path)) {
throw new RuntimeException("$path does not exist.");
}
$filtered += ['.', '..'];
$dirs = [];
$d = dir($path);
while (($entry = $d->read()) !== false) {
if (is_dir("$path/$entry") && !in_array($entry, $filtered)) {
$dirs[] = $entry;
if ($recursive) {
$newDirs = getDirs("$path/$entry");
foreach ($newDirs as $newDir) {
$dirs[] = "$entry/$newDir";
}
}
}
}
return $dirs;
}
于 2020-04-12T08:47:16.580 回答
2
<?php
/*this will do what you asked for, it only returns the subdirectory names in a given
path, and you can make hyperlinks and use them:
*/
$yourStartingPath = "photos\\";
$iterator = new RecursiveIteratorIterator(
new RecursiveDirectoryIterator($yourStartingPath),
RecursiveIteratorIterator::SELF_FIRST);
foreach($iterator as $file) {
if($file->isDir()) {
$path = strtoupper($file->getRealpath()) ;
$path2 = PHP_EOL;
$path3 = $path.$path2;
$result = end(explode('/', $path3));
echo "<br />". basename($result );
}
}
/* best regards,
Sanaan Barzinji
Erbil
*/
?>
于 2012-08-04T07:41:53.787 回答
1
正确的方式
/**
* Get all of the directories within a given directory.
*
* @param string $directory
* @return array
*/
function directories($directory)
{
$glob = glob($directory . '/*');
if($glob === false)
{
return array();
}
return array_filter($glob, function($dir) {
return is_dir($dir);
});
}
灵感来自 Laravel
于 2016-06-01T22:55:42.010 回答
1
以下递归函数返回一个包含子目录完整列表的数组
function getSubDirectories($dir)
{
$subDir = array();
$directories = array_filter(glob($dir), 'is_dir');
$subDir = array_merge($subDir, $directories);
foreach ($directories as $directory) $subDir = array_merge($subDir, getSubDirectories($directory.'/*'));
return $subDir;
}
来源：https ://www.lucidar.me/en/web-dev/how-to-get-subdirectories-in-php/
于 2018-09-21T04:33:04.730 回答
1
这是一个班轮代码：
$sub_directories = array_map('basename', glob($directory_path . '/*', GLOB_ONLYDIR));
于 2019-07-25T11:48:53.663 回答
0
您可以使用 glob() 函数来执行此操作。
这是一些关于它的文档：http: //php.net/manual/en/function.glob.php
于 2015-04-16T14:49:42.743 回答
0
递归查找所有 PHP 文件。逻辑应该足够简单以进行调整，并且旨在通过避免函数调用来更快（更）。
function get_all_php_files($directory) {
$directory_stack = array($directory);
$ignored_filename = array(
'.git' => true,
'.svn' => true,
'.hg' => true,
'index.php' => true,
);
$file_list = array();
while ($directory_stack) {
$current_directory = array_shift($directory_stack);
$files = scandir($current_directory);
foreach ($files as $filename) {
// Skip all files/directories with:
// - A starting '.'
// - A starting '_'
// - Ignore 'index.php' files
$pathname = $current_directory . DIRECTORY_SEPARATOR . $filename;
if (isset($filename[0]) && (
$filename[0] === '.' ||
$filename[0] === '_' ||
isset($ignored_filename[$filename])
))
{
continue;
}
else if (is_dir($pathname) === TRUE) {
$directory_stack[] = $pathname;
} else if (pathinfo($pathname, PATHINFO_EXTENSION) === 'php') {
$file_list[] = $pathname;
}
}
}
return $file_list;
}
于 2016-12-07T09:54:11.777 回答
-1
查找指定目录下的所有文件和文件夹。
function scanDirAndSubdir($dir, &$fullDir = array()){
$currentDir = scandir($dir);
foreach ($currentDir as $key => $val) {
$realpath = realpath($dir . DIRECTORY_SEPARATOR . $val);
if (!is_dir($realpath) && $filename != "." && $filename != "..") {
scanDirAndSubdir($realpath, $fullDir);
$fullDir[] = $realpath;
}
}
return $fullDir;
}
var_dump(scanDirAndSubdir('C:/web2.0/'));
样本 ：
array (size=4)
0 => string 'C:/web2.0/config/' (length=17)
1 => string 'C:/web2.0/js/' (length=13)
2 => string 'C:/web2.0/mydir/' (length=16)
3 => string 'C:/web2.0/myfile/' (length=17)
于 2015-11-05T17:09:26.140 回答
-1
如果您正在寻找递归目录列表解决方案。使用下面的代码，希望对您有所帮助。
<?php
/**
* Function for recursive directory file list search as an array.
*
* @param mixed $dir Main Directory Path.
*
* @return array
*/
function listFolderFiles($dir)
{
$fileInfo = scandir($dir);
$allFileLists = [];
foreach ($fileInfo as $folder) {
if ($folder !== '.' && $folder !== '..') {
if (is_dir($dir . DIRECTORY_SEPARATOR . $folder) === true) {
$allFileLists[$folder . '/'] = listFolderFiles($dir . DIRECTORY_SEPARATOR . $folder);
} else {
$allFileLists[$folder] = $folder;
}
}
}
return $allFileLists;
}//end listFolderFiles()
$dir = listFolderFiles('your searching directory path ex:-F:\xampp\htdocs\abc');
echo '<pre>';
print_r($dir);
echo '</pre>'
?>
于 2017-05-27T18:05:24.167 回答