is_uploaded_file

(PHP 3 >= 3.0.17, PHP 4 >= 4.0.3, PHP 5)

is_uploaded_file -- Определяет, был ли файл загружен при помощи HTTP POST

Описание

bool is_uploaded_file ( string filename )

Возвращает TRUE, если файл filename был загружен при помощи HTTP POST. Это полезно, чтобы убедиться в том, что злонамеренный пользователь не пытается обмануть скрипт так, чтобы он работал с файлами, с которыми работать не должен -- к примеру, /etc/passwd.

Такие проверки особенно полезны, если существует вероятность того, что операции над файлом могут показать его содержимое пользователю или даже другим пользователям той же системы.

Для правильной работы, функции is_uploaded_file() нужен аргумент вида $_FILES['userfile']['tmp_name'], - имя закачиваемого файла на клиентской машине $_FILES['userfile']['name'] не подходит.

Пример 1. Пример использования функции is_uploaded_file()

<?php

if (is_uploaded_file($_FILES['userfile']['tmp_name'])) {
   echo "File ". $_FILES['userfile']['name'] ." uploaded successfully.\n";
   echo "Displaying contents\n";
   readfile($_FILES['userfile']['tmp_name']);
} else {
   echo "Possible file upload attack: ";
   echo "filename '". $_FILES['userfile']['tmp_name'] . "'.";
}

?>

Функция is_uploaded_file() доступна только начиная с версии PHP 3.0.16 и PHP 4.0.2 для третьей и четвёртой линеек соответственно. Если вам необходимо использовать более ранние версии, вы можете использовать такую функцию, чтобы обезопасить себя:

Замечание: Следующий пример не работает в версиях PHP выше 4.0.2. Он полагается на внутреннюю функциональность PHP, которая изменилась после этой версии.

Пример 2. Пример использования функции is_uploaded_file()

<?php
/* Самодельная проверка на то, является ли файл загруженным. */
function is_uploaded_file($filename)
{
   if (!$tmp_file = get_cfg_var('upload_tmp_dir')) {
       $tmp_file = dirname(tempnam('', ''));
   }
   $tmp_file .= '/' . basename($filename);
   /* У пользователя может быть завершающий слеш в php.ini... */
   return (ereg_replace('/+', '/', $tmp_file) == $filename);
}

/* Вот пример использования, так как функция move_uploaded_file()
 * также отсутствует в ранних версиях: */
if (is_uploaded_file($HTTP_POST_FILES['userfile'])) {
   copy($HTTP_POST_FILES['userfile'], "/place/to/put/uploaded/file");
} else {
   echo "Возможна атака загрузки файла: имя файла - '$HTTP_POST_FILES[userfile]'.";
}
?>

См. также описание функции move_uploaded_file() и раздел Загрузка файлов на сервер для получения простых примеров использования.



is_uploaded_file
tejas at kaushalam dot com
28-Mar-2006 11:28
<FORM ENCTYPE="multipart/form-data" ACTION="_URL_" METHOD=POST>
Upload this file: <INPUT NAME="userfile" TYPE="file">
<INPUT TYPE="submit" VALUE="Send File"></FORM>

This html code will display a text area with file upload button. The form tag is bit different than the normal form tag used ( see the encrypte =).

Now let us go to the php part to handle the uploaded file. We have In PHP 3, the following variables will be defined within the destination script upon a successful upload, assuming that register_globals is turned on in php3.ini. If track_vars is turned on, they will also be available in PHP 3 within the global

array $HTTP_POST_VARS. Note that the following variable names assume the use of the file upload name 'userfile', as used in the example above:

   * $userfile - The temporary filename in which the uploaded file was stored on the server machine.
   * $userfile_name - The original name or path of the file on the sender's system.
   * $userfile_size - The size of the uploaded file in bytes.
   * $userfile_type - The mime type of the file if the browser provided this information. An example would be "image/gif".

With all these info we will make our script ready to handle the files.
Let us check the file size and we will not allow file size more than 250 KB to get uploaded to our server. Here we are using a flag $file_upload to false for processing the file upload.

if ($userfile_size >250000){$msg=$msg."Your uploaded file size is more than 250KB so please reduce the file size and then upload. Visit the help page to know how to reduce the file size.<BR>";
$file_upload="false";}

Now let us check the file extension and only jpg or gif file pictures we will allow into our server. We will check this by using $userfile_type extension

if (!($userfile_type =="image/pjpeg" OR $userfile_type=="image/gif")){$msg=$msg."Your uploaded file must be of JPG or GIF. Other file types are not allowed<BR>";
$file_upload="false";}
We will limit ourselves to these two type checks and if we find our $file_upload variable is not "false" then we can upload the file. This part is very simple in PHP and can be done in one step. Before that let us decide the file directory name where we will be placing the file. $add is the path with the file name relative to the script running this code where the uploaded file will be stored.

$add="upload/$userfile_name"; // the path with the file name where the file will be stored, upload is the directory name.

Now let us copy the file to server. The command move_uploaded_file will does that job for us and if the action is successful then it will return true.

if(move_uploaded_file ($userfile, $add)){
// do your coding here to give a thanks message or any other thing.
}else{echo "Failed to upload file Contact Site admin to fix the problem";}
Thats all... the file is placed in our directory (name: upload)
YLearn
10-Oct-2005 10:42
Just looked at what I posted again and found several mistakes of the major and minor sort.  That's what I get for posting before I finish my coffee.  This should work better (i.e. should work in the first place):

<?php
   default: //a default error, just in case!  :)
       echo "There was a problem with your upload.";
       $err_msg = "Unrecognized file POST error: ".$HTTP_POST_FILES['userfile']['error'];
       if (!(strpos($err_msg, "\n") === false)) {
           $err_lines = explode("\n", $err_msg);
           foreach ($err_lines as $msg) {
               error_log($msg, 0);
           }
       } else {
           error_log($err_msg, 0);
       }
       break;
?>
YLearn
10-Oct-2005 09:41
Regarding topcat's suggested change, I am split on doing that.  I don't like showing users errors that may give them more information than they should have (or show that I haven't provided for that particular error).  But I want to know when there are errors that fall to the default case so I can fix my code.  What I will typically do is write them to the error log something like this modification to metaltoad's post (takes into account the possibility of multi-line errors which error_log doesn't handle well):

<?php
   default: //a default error, just in case!  :)
       echo "There was a problem with your upload.";
       $err_msg = "Unrecognized file POST error: ".$HTTP_POST_FILES['userfile']['error'];
       if ((strpos($err_msg, "\n") === 0) {
           $err_lines = explode("\n", $err_msg);
           foreach ($err_lines as $msg) {
               error_log($msg, 0);
           }
       } else {
           error_log($err_msg, 0)
       }
       break;
?>
juk
18-Sep-2005 03:26
If your $_FILES and $_POST are empty, this can be due to
- the limit set by post_max_size in php.ini
- the limit set by upload_max_filesize in php.ini

Unfortunately the first limit is not reported back as an error code in $_FILES['error'].
topcat
14-Jul-2005 04:56
Just a little tip to info at metaltoad's comment:
It's good practice to print error code when it can't be recognized:

   default: //print the error code
     echo "Unrecognized error code: ".$HTTP_POST_FILES['userfile']['error'];
     break;
23-Apr-2005 12:29
make use u got the enctype="multipart/form-data" in ur form tag otrherwise nothing works... took me two hours to find that out.......
beer UNDRSCR nomaed AT hotmail DOT com
15-Apr-2005 04:21
Regarding the comment of info at metaltoad dot net
@ 19-Feb-2003 04:03

<?php
// ... yada yada yada...
preg_match("/.exe$|.com$|.bat$|.zip$|.doc$|.txt$/i", $HTTP_POST_FILES['userfile']['name']))
// ... yada yada yada...
?>

This will not work. It will, but not correctly.
You shuld escape the . (dot) for the preg function,
and escape the $ (dollar) sign for PHP, or use
single-quoted string...

The syntax should be (much shorter and neater):

<?php
// ... yada yada yada...
preg_match('/\\.(exe|com|bat|zip|doc|txt)$/i', $_FILES['userfile']['name']))
// ... yada yada yada...
?>
lots2learn at gmail dot com
06-Feb-2005 09:13
if files are not getting uploaded and $_FILE array is empty ..and your code looks fine..then check php.ini file..the file_uploads option should be turned 'On' to allow file uploads. Turn it on and restart apache to have effect .
Gordon Luk
04-Oct-2004 03:55
If the $_FILES array suddenly goes mysteriously empty, even though your form seems correct, you should check the disk space available for your temporary folder partition. In my installation, all file uploads failed without warning. After much gnashing of teeth, I tried freeing up additional space, after which file uploads suddenly worked again.
vbudov_yahoo.com
21-May-2004 04:58
Before moving the file in to place it's also a good idea to check if file with the same name already exists on the server.
If file exists then create unique name for the new file.

$num=1;
while (file_exists($destination)){   
   $num++; // if previous file name existed then thy another number+_+filename                                                                                                     
   $file_name = $num."_".$_FILES['userfile']['name'];
   $destination = $uploadpath.$file_name;
}                                                                                                                                 
move_uploaded_file( $source, $destination );
phpnetmark at nunswithguns dot co dot uk
16-Apr-2003 10:53
If you are importing the uploaded file into a BLOB field in a mysql database and you are using LOAD_FILE() sql statement then be aware that  mysql checks max-allowed-packet mysql variable.

- if the size of the binary file LOAD_FILE is importing is bigger than
max-allowed-packet size then it LOAD_FILE will return null.

You can specify max-allowed-packet size in the call to mysqld eg:

./bin/safe_mysqld --user=mysql --max-allowed-packet=16M &

Full info on this variable is available here:
http://www.mysql.com/doc/en/Packet_too_large.html
info at metaltoad dot net
19-Feb-2003 01:03
As of PHP 4.2.0, rather than automatically assuming a failed file uploaded is a file attack, you can use the error code associated with the file upload to check and see why the upload failed.  This error code is stored in the userfile array (ex: $HTTP_POST_FILES['userfile']['error']).

Here's an example of a switch:

if (is_uploaded_file($userfile)) {
 
  //include code to copy tmp file to final location here...
 
}else{
  switch($HTTP_POST_FILES['userfile']['error']){
   case 0: //no error; possible file attack!
     echo "There was a problem with your upload.";
     break;
   case 1: //uploaded file exceeds the upload_max_filesize directive in php.ini
     echo "The file you are trying to upload is too big.";
     break;
   case 2: //uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the html form
     echo "The file you are trying to upload is too big.";
     break;
   case 3: //uploaded file was only partially uploaded
     echo "The file you are trying upload was only partially uploaded.";
     break;
   case 4: //no file was uploaded
     echo "You must select an image for upload.";
     break;
   default: //a default error, just in case!  :)
     echo "There was a problem with your upload.";
     break;
}

Additionally, by testing the 'name' element of the file upload array, you can filter out unwanted file types (.exe, .zip, .bat, etc).  Here's an example of a filter that can be added before testing to see if the file was uploaded:

//rejects all .exe, .com, .bat, .zip, .doc and .txt files
if(preg_match("/.exe$|.com$|.bat$|.zip$|.doc$|.txt$/i", $HTTP_POST_FILES['userfile']['name'])){
  exit("You cannot upload this type of file.");
}

//if file is not rejected by the filter, continue normally
if (is_uploaded_file($userfile)) {

/*rest of code*/
itadmin at itmusicweb dot co dot uk
28-Nov-2002 06:11
The example brought out does not work as supposed to:

function is_uploaded_file($filename) {
   if (!$tmp_file = get_cfg_var('upload_tmp_dir')) {
       $tmp_file = dirname(tempnam('', ''));
   }
   $tmp_file .= '/' . basename($filename);
   /* User might have trailing slash in php.ini... */
   return (ereg_replace('/+', '/', $tmp_file) == $filename);
}

It works only with files under ....4 or 5 kb, other files automatically get the size of 0 bytes. So something must be wrong here. Built-in is_uploaded_file() works good.
troels at NO dot SPAM dot webcode dot dk
14-Oct-2002 12:44
to get the example to work on windows, youll have to add a line, that replaces backslashes with slashes. eg.: $filename = str_replace ("\\", "/", $filename);

also, as someone mentioned, globalizing $HTTP_POST_FILES is a good idea ...

<pre>
/* Userland test for uploaded file. */
function is_uploaded_file($filename)
{
   global $HTTP_POST_FILES;
   if (!$tmp_file = get_cfg_var("upload_tmp_dir")) {
       $tmp_file = dirname(tempnam("", ""));
   }
   $tmp_file .= "/" . basename($filename);
   /* User might have trailing slash in php.ini... */
   // fix for win platform
   $filename = str_replace ("\\", "/", $filename);
   return (ereg_replace("/+", "/", $tmp_file) == $filename);
}
</pre>
r3gan at hotmail dot com
17-Jun-2002 12:24
Remeber, if using $HTTP_POST_FILES inside a function and it doesn't seem to work, try globalizing the array:

function UploadFile() {

   global $HTTP_POST_FILES;

   // rest of your code here

}  // end UploadFile

<is_readableis_writable>
 Last updated: Tue, 15 Nov 2005