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up vote 1 down vote favorite

Hi thx to tutorials and own hinking I wrote this code

if(isset($_POST['upload'])) {

$allowed_filetypes = array('.jpg','.jpeg','.png','.gif');
$max_filesize = 10485760;
$upload_path = 'uploads/';
$description = $_POST['imgdesc'];

$filename = $_FILES['userfile']['name'];
$ext = substr($filename, strpos($filename,'.'), strlen($filename)-1);

if(!in_array($ext,$allowed_filetypes))
  die('The file you attempted to upload is not allowed.');

if(filesize($_FILES['userfile']['tmp_name']) > $max_filesize)
  die('The file you attempted to upload is too large.');

if(!is_writable($upload_path))
  die('You cannot upload to the specified directory, please CHMOD it to 777.');

if(move_uploaded_file($_FILES['userfile']['tmp_name'],$upload_path . $filename)) {
   $query = "INSERT INTO uploads (name, description) VALUES ($filename, $description)"; 
   mysql_query($query);

echo 'Your file upload was successful!';


} else {
     echo 'There was an error during the file upload.  Please try again.';
}
}

I'm getting this error;

Notice: Undefined index: userfile in C:\WebServer\htdocs\PicSide\admin\addimage.php on line 16

How to fix it anyone know? And is this correct to upload o server?

share|improve this question
    
can you post your html form, we need to see that as well? also notices are not exactly errors, errors stops the rest of the script from running whereas notices - depending on the server environment - do not. –  ncremins Dec 28 '12 at 23:11
    
Probably the file input element name is not userfiles –  Sinan Dec 28 '12 at 23:12
    
1st Thank you for reply, I noticed I didn't name my input by "userfile" dummy me, but otherwise is this code correct? –  Piggie Dec 28 '12 at 23:13
    
It should be correct –  Вукашин Пиштељић Dec 28 '12 at 23:14
    
@creminsn, Notices are not errors? I wonder why it is an option in PHP error reporting? =oP It is an error, just a less severe error. Notices should be fixed as having them means the code is not properly being done to account for variables not existing, which does not make robust code. –  cryptic ツ Dec 28 '12 at 23:18

3 Answers 3

up vote 4 down vote accepted

Basically, what your error means is that you don't have the following (or somewhat similar) element in your form:

<input type="file" name="userfile" />

The important part in this is the actual name="userfile" which specifies the key in the $_FILES array.

share|improve this answer
    
Thanks! I actually found it out 1min after posting this, I noticed to late, sorry for unnecessary post –  Piggie Dec 28 '12 at 23:22
2  
@Piggie, I personally would mark ninetwozero's as the solution. –  cryptic ツ Dec 28 '12 at 23:31
up vote 0 down vote

And don't forget to concat:

$query = "INSERT INTO uploads (name, description) VALUES ($filename, $description)";

Should be:

$query = "INSERT INTO uploads (name, description) VALUES (".$filename.", ".$description.")";
share|improve this answer
up vote -2 down vote 
<?php
error_reporting(0);
?>
share|improve this answer
    
Funny. Great solution –  chappjc Jan 24 at 6:37

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