PHP - Manual: is_dir

2025-11-01

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is_dir

(PHP 4, PHP 5, PHP 7, PHP 8)

is_dir — 判断给定文件名是否是一个目录

说明

is_dir(string $filename): bool

判断给定文件名是否是一个目录。

参数

filename: 文件路径，如果 filename 是相对文件名，会相对于当前工作目录进行检查。如果 filename 是符号链接或者硬链接，然后解析链接并检查。如果启用了 open_basedir，则会应用更多限制。

返回值

如果文件名存在，并且是个目录，返回 true，否则返回false。

错误／异常

失败时抛出 E_WARNING 警告。

示例

示例 #1 is_dir() 例子

<?php
var_dump(is_dir('a_file.txt'));
var_dump(is_dir('bogus_dir/abc'));

var_dump(is_dir('..')); //one dir up
?>

以上示例会输出：

bool(false)
bool(false)
bool(true)

注释

注意: 此函数的结果会被缓存。参见 clearstatcache() 以获得更多细节。

小技巧

自 PHP 5.0.0 起, 此函数也用于某些 URL 包装器。请参见支持的协议和封装协议以获得支持 stat() 系列函数功能的包装器列表。

参见

chdir() - 改变目录
dir() - 返回一个 Directory 类实例
opendir() - 打开目录句柄
is_file() - 判断给定文件名是否为一个正常的文件
is_link() - 判断给定文件名是否为一个符号连接

发现了问题？

了解如何改进此页面 • 提交拉取请求 • 报告一个错误

＋添加备注

用户贡献的备注 20 notes

down

digitalaudiorock at gmail dot com ¶

14 years ago

Just a note for anyone who encounters is_dir() returning false on CIFS mount points or directories within those mount points on 2.6.31 and newer kernels: Apparently in new kernels they've started using the CIFS serverino option by default.  With Windows shares this causes huge inode numbers and which apparently can cause is_dir() to return false.  Adding the noserverino option to the CIFS mount will prevent this.  This may only occur on 32 systems but I don't have a 64 bit install to test against.

down

thomas at thomasnoest dot nl ¶

9 years ago

Note that on Linux is_dir returns FALSE if a parent directory does not have +x (executable) set for the php process.

down

sly at noiretblanc dot org ¶

20 years ago

This is the "is_dir" function I use to solve the problems :

function Another_is_dir ($file)
{
    if ((fileperms("$file") & 0x4000) == 0x4000)
        return TRUE;
    else
        return FALSE;
}

or, more simple :

function Another_is_dir ($file)
{ 
return ((fileperms("$file") & 0x4000) == 0x4000);
}

I can't remember where it comes from, but it works fine.

down

jonlulf at gmail dot com ¶

9 years ago

My solution to the problem that you must include the full path to make "is_dir" work properly as a complete example:

    <? // findfiles.php  -  what is in directory "videoarchive"
    $dir    = 'images/videoarchive/'; // path from top
    $files = scandir($dir);
    $files_n = count($files);


    $i=0;
    while($i<=$files_n){
        // "is_dir" only works from top directory, so append the $dir before the file
        if (is_dir($dir.'/'.$files[$i])){   
            $MyFileType[$i] = "D" ; // D for Directory
        } else{
            $MyFileType[$i] = "F" ; // F for File
        }
        // print itemNo, itemType(D/F) and itemname
        echo '<br>'.$i.'. '. $MyFileType[$i].'. ' .$files[$i] ;
        $i++;
    }
    ?>

down

niceuser at live dot com ¶

11 years ago

PITFALL in sub dir processing

After struggeling with a sub-dir processing (some subdirs were skipped) AND reading the posts, I realized that virutally no-one clearly told what were wrong.

The common traverse dir code was:
-----------------------------------------

opendir("myphotos"); // Top dir to process from (example)

while (false !== ($fname = readdir($h_dir))) { // process current dir (read a directory entry)

   if ($fname{0} == '.') continue; // skip dirs . and .. by first char test

   if (is_dir($fname)) call_own_subdir_process;  // process this subdir by calling a routine

   }

PROBLEM IS :

The "is_dir()" must have the FULL PATH or it will skip some dirs. So the above code need to INSERT THE PATH before the filename. This would give this change in above...

   if (is_dir("myphotos\" . $fname)) call_own_subdir_process;  // skip subdirs

The pitfall really was, that without full path some subdirs were found...hope this clears all up

down

vstoykov at consultant dot bg ¶

15 years ago

When trying (no 'pear') to enumerate mounted drives on a win32  platform (Win XP SP3, Apache/2.2.11, PHP/5.2.9), I used:

<?php
function echo_win_drives() {

  for($c='A'; $c<='Z'; $c++) 
    if(is_dir($c . ':'))
      echo $c . ': '; 
}
?>

which yielded:
A: C: D: E: F: G: H: I:

down

Eric ¶

18 years ago

Ah ha!  Maybe this is a bug, or limitation to be more precise, of php. See http://bugs.php.net/bug.php?id=27792

A workaround is posted on the page (above) and seems to work for me:

function is_dir_LFS($path){
  return (('d'==substr(exec("ls -dl '$path'"),0,1))?(true):(false));
}

PS: I'm using PHP 4.3.10-16, posts report this problem up to 5.0

down

danr at cvisual dot com dot com ¶

17 years ago

Running PHP 5.2.0 on Apache Windows, I had a problem (likely the same one as described by others) where is_dir returned a False for directories with certain permissions even though they were accessible.

Strangely, I was able to overcome the problem with a more complete path. For example, this only displays "Works" on subdirectories with particular permissions (in this directory about 1 out of 3):

$d = opendir("./albums/mydir");
while(false !== ($f = readdir($d))) {
    echo "<hr />";
        if(is_dir($f)) {
            echo "<b>Works:" . $f . "</b>";
        }
}

However, this works properly for all directories:

$d = opendir("./albums/mydir");
while(false !== ($f = readdir($d))) {
    echo "<hr />";
        $dName = "./albums/mydir/" . $f;
        if(is_dir($dName)) {
            echo "<b>Works:" . $dName . "</b>";
        }
}

I don't understand the hit-and-miss of the first code, but maybe the second code can help others having this problem.

down

Anonymous ¶

17 years ago

One note regarding checking for empty directories :
>>echo (count(glob("$dir/*")) === 0) ? 'Empty' : 'Not empty';
This does not work correctly on Linux.
The '.' and '..' will always be returned even if no files are present in the directory.

down

musicmaster ¶

1 year ago

is_dir() doesn't work with a directory that has the name "0" (a single zero). Neither does realpath().

This is my experience with PHP 7.4.

down

jerome dot pros at gmail dot com ¶

6 years ago

Note that is_dir() also works with ftp://.

For example :

<?php 
if(is_dir('ftp://user:pass@host/www/path/to/your/folder')) {
// Your code.
}
?>

But note that if the connexion fails due to invalide credentials, this will consider that the folder doesn't exist and will return FALSE.

down

Bjrn K. ¶

8 years ago

Note that this functions follows symbolic links. It will return true if the file is actually a symlink that points to a directory.

An example:
<php
symlink(".", "testlink");
var_dump(is_dir("testlink"));
unlink("testlink");
?>

Prints out:
bool(true)

(Windows Note: Under recent versions of Windows you can set symlinks as long as you're administrator, but you cannot remove directory symlinks with "unlink()", you will have to use "rmdir testlink" from the shell to get rid of it.)

down

gecko4 at gmail dot com ¶

18 years ago

Here is another way to test if a directory is empty, which I think is much simpler than those posted below:

<?php
$dir = 'directory';
echo (count(glob("$dir/*")) === 0) ? 'Empty' : 'Not empty';
?>

down

puremango dot co dot uk at gmail dot com ¶

20 years ago

this function bypasses open_basedir restrictions.
<?
function my_is_dir($dir)
{
    // bypasses open_basedir restrictions of is_dir and fileperms
    $tmp_cmd = `ls -dl $dir`;
    $dir_flag = $tmp_cmd[0];
    if($dir_flag!="d")
    {
        // not d; use next char (first char might be 's' and is still directory)
        $dir_flag = $tmp_cmd[1];
    }
    return ($dir_flag=="d");
}
?>

example:
<?
....
echo is_dir("/somewhere/i/dont/have/access/to");
?>
output:
Warning: open_basedir restriction in effect

<?
....
echo my_is_dir("/somewhere/i/dont/have/access/to");
?>
output:
true (or false, depending whether it is or not...)

---
visit puremango.co.uk for other such wonders

down

-1

citizenmarco at gmail dot com ¶

8 years ago

Note that there quite a few articles on the net that imply that commands like is_dir, opendir, readdir cannot read paths with spaces.

On a linux box, THAT is not an issue. 

Sample test code;

 $dir = "Images/Soma ALbum Name with spaces";


echo $dir."<br/>";

// Open a directory, and read its contents
if (is_dir($dir)){
  echo $dir."<br/>"; // will not appear if above fails
    if ($dh = opendir($dir)){
      echo $dir."<br/>"; // will not appear if above fails
      while (($file = readdir($dh)) !== false){
        echo "filename:" . $file . "<br>";
        echo $dir."<br/>"; // will not appear if above fails
      }
      closedir($dh);
    }
}

down

-2

Victor ¶

11 years ago

If you are using Mac, or others systems that store information about the directory layout and etc, the function:

   function empty_dir($dir) {
        if (($files = @scandir($dir)) && count($files) <= 3)
            return true;
        else
            return false;
    }

Must have the count($files) comparing with the number of hidden files!

For example, I'm using Mac and the empty directory shows me three files: ".", ".." and ".DS_Store", so if I am planning to put the website online on my Mac, I've to count in the ".DS_Store" file!

down

-3

jasoneisen at gee mail ¶

16 years ago

An even better (PHP 5 only) alternative to "Davy Defaud's function": 

<?php
function is_empty_dir($dir)
{
    if (($files = @scandir($dir)) && count($files) <= 2) {
        return true;
    }
    return false;
}
?>

NOTE: you should obviously be checking beforehand if $dir is actually a directory, and that it is readable, as only relying on this you would assume that in both cases you have a non-empty readable directory.

down

-3

Anonymous ¶

20 years ago

Unfortunately, the function posted by p dot marzec at bold-sg dot pl does not work.
The corrected version is:

// returns true if folder is empty or not existing
// false if folde is full

function is_empty_folder($dir) {
if (is_dir($dir)) {
   $dl=opendir($dir);
   if ($dl) {
       while($name = readdir($dl)) {
   if (!is_dir("$dir/$name")) { //<--- corrected here
       return false;
       break;
       }
   }
       closedir($dl);
       }
   return true;
   } else return true;
}

down

-5

Btx ¶

16 years ago

<?php

public static function isEmptyDir($dir){

     return (($files = @scandir($dir)) && count($files) <= 2);

}

?>



better ;)

down

-5

tibard at gmail dot com ¶

20 years ago

use this function to get all files inside a directory (including subdirectories)

<?php
function scan_Dir($dir) {
$arrfiles = array();
    if (is_dir($dir)) {
        if ($handle = opendir($dir)) {
chdir($dir);
            while (false !== ($file = readdir($handle))) { 
                if ($file != "." && $file != "..") { 
                    if (is_dir($file)) { 
$arr = scan_Dir($file);
                        foreach ($arr as $value) {
$arrfiles[] = $dir."/".$value;
                        }
                    } else {
$arrfiles[] = $dir."/".$file;
                    }
                }
            }
chdir("../");
        }
closedir($handle);
    }
    return $arrfiles;
}

?>

＋添加备注

官方地址：https://www.php.net/manual/en/function.is-dir.php

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PHP官方手册 - 互联网笔记

is_dir

说明

参数

返回值

错误／异常

示例

注释

参见

发现了问题？

用户贡献的备注 20 notes