Read this section of the manual... http://www.php.net/manual/en/features.file-upload.php
Assuming your form input looks like this: <input type=file name='userfile'> then you'll want to use: $_FILES['userfile']['name'] The original name of the file on the client machine. $_FILES['userfile']['type'] The mime type of the file, if the browser provided this information. An example would be "image/gif". $_FILES['userfile']['size'] The size, in bytes, of the uploaded file. $_FILES['userfile']['tmp_name'] The temporary filename of the file in which the uploaded file was stored on the server. -p On Wed, 22 May 2002, Ragnar wrote: > Hi, > > im a newbie to php and are trying to make a small upload script. So i have > read about the variables $userfile and $userfile_name. When i try to run > the > script i get the following error: > > Notice: Undefined variable: userfile > Notice: Undefined variable: userfile_name > > Some of the code: > > if ( is_uploaded_file ( $_FILES [$userfile] [$userfile_name] ) ) > > Why cant i use the variables? > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
