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.
is_dir
(PHP 4, PHP 5, PHP 7, PHP 8)
is_dir — Определяет, является ли имя файла директорией
Список параметров
-
filename -
Путь к файлу. Если
filenameявляется относительным именем, он будет проверяться относительно текущей рабочей директории. Еслиfilenameявляется символической или жёсткой ссылкой, то ссылка будет раскрыта и проверена. При включённом open_basedir могут применяться дальнейшие ограничения.
Возвращаемые значения
Возвращает true, если файл существует и является директорией,
иначе возвращается false.
Ошибки
В случае неудачного завершения работы генерируется ошибка уровня E_WARNING.
Примеры
Пример #1 Пример использования функции is_dir()
<?php
var_dump(is_dir('a_file.txt'));
var_dump(is_dir('bogus_dir/abc'));
var_dump(is_dir('..')); // на одну директорию выше
?>Результат выполнения приведённого примера:
bool(false) bool(false) bool(true)
Примечания
Замечание: Результаты этой функции кешируются. Более подробную информацию смотрите в разделе clearstatcache().
Подсказка
Начиная с PHP 5.0.0, эта функция также может быть использована с некоторыми обёртками url. Список обёрток, поддерживаемых семейством функций stat(), смотрите в разделе Поддерживаемые протоколы и обёртки.
add a note
User Contributed Notes 21 notes
digitalaudiorock at gmail dot com ¶
14 years ago
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.
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] ;
$i++;
}
?>
sly at noiretblanc dot org ¶
20 years ago
This is the "is_dir" function I use to solve the problems :
function Another_is_dir ($file)
{
if ((fileperms("$file") & 0x4000) == 0x4000)
return TRUE;
else
return FALSE;
}
or, more simple :
function Another_is_dir ($file)
{
return ((fileperms("$file") & 0x4000) == 0x4000);
}
I can't remember where it comes from, but it works fine.
jerome dot pros at gmail dot com ¶
5 years ago
Note that is_dir() also works with ftp://.
For example :
<?php
if(is_dir('ftp://user:pass@host/www/path/to/your/folder')) {
// Your code.
}
?>
But note that if the connexion fails due to invalide credentials, this will consider that the folder doesn't exist and will return FALSE.
vstoykov at consultant dot bg ¶
15 years ago
When trying (no 'pear') to enumerate mounted drives on a win32 platform (Win XP SP3, Apache/2.2.11, PHP/5.2.9), I used:
<?php
function echo_win_drives() {
for($c='A'; $c<='Z'; $c++)
if(is_dir($c . ':'))
echo $c . ': ';
}
?>
which yielded:
A: C: D: E: F: G: H: I:
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:" . $dName . "</b>";
}
}
I don't understand the hit-and-miss of the first code, but maybe the second code can help others having this problem.
niceuser at live dot com ¶
11 years ago
PITFALL in sub dir processing
After struggeling with a sub-dir processing (some subdirs were skipped) AND reading the posts, I realized that virutally no-one clearly told what were wrong.
The common traverse dir code was:
-----------------------------------------
opendir("myphotos"); // Top dir to process from (example)
while (false !== ($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\" . $fname)) call_own_subdir_process; // skip subdirs
The pitfall really was, that without full path some subdirs were found...hope this clears all up
Anonymous ¶
16 years ago
One note regarding checking for empty directories :
>>echo (count(glob("$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.
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.
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.)
gecko4 at gmail dot com ¶
17 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';
?>
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
puremango dot co dot uk at gmail dot com ¶
19 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
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);
}
}
tibard at gmail dot com ¶
19 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 $arrfiles;
}
?>
jasoneisen at gee mail ¶
16 years ago
An even better (PHP 5 only) alternative to "Davy Defaud's function":
<?php
function is_empty_dir($dir)
{
if (($files = @scandir($dir)) && count($files) <= 2) {
return true;
}
return false;
}
?>
NOTE: you should obviously be checking beforehand if $dir is actually a directory, and that it is readable, as only relying on this you would assume that in both cases you have a non-empty readable directory.
Btx ¶
16 years ago
<?php
public static function isEmptyDir($dir){
return (($files = @scandir($dir)) && count($files) <= 2);
}
?>
better ;)
Victor ¶
11 years ago
If you are using Mac, or others systems that store information about the directory layout and etc, the function:
function empty_dir($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!
alanguir at detroitchamber dot com ¶
15 years ago
When I run a scandir I always run a simple filter to account for file system artifacts (especially from a simple ftp folder drop) and the "." ".." that shows up in every directory:
<?php
if (is_dir($folder){
$contents = scandir($folder);
$bad = array(".", "..", ".DS_Store", "_notes", "Thumbs.db");
$files = array_diff($contents, $bad);
}
?>
Anonymous ¶
19 years ago
Unfortunately, the function posted by p dot marzec at bold-sg dot pl does not work.
The corrected version is:
// returns true if folder is empty or not existing
// false if folde is full
function is_empty_folder($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;
}
