Bonzol

I have the following code [code=php:0] <?php $uploaddir = 'ftp://username:password@000.000.000.00/public_ftp/'; $uploadfile = $uploaddir . basename($_FILES['userfile']['name']); echo '<pre>'; if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) { echo "File is valid, and was successfully uploaded.\n"; } else { echo "Possible file upload attack!\n"; } echo 'Here is some more debugging info:'; print_r($_FILES); print "</pre>"; ?> [/code] The problem is, this appears to work, and the file appears in the ftp server, however the file has no size, its almost as though all that has happend is the file name has been copied and nothing else,,anyone got any suggestions on how to copy the file correctly? Thanx in advanced

Hey there, I am trying to create a php file upload page and have run in to a problem. going through some tutorials I have come up with the following code my form <form enctype="multipart/form-data" action="uptest.php" method="POST"> <input type="hidden" name="MAX_FILE_SIZE" value="30000" /> Send this file: <input name="userfile" type="file" /> <input type="submit" value="Send File" /> </form> and the php code, ftp URL is not exact for privacy reasons [code=php:0] <?php $uploaddir = 'ftp://username:password@000.000.0.0/public_ftp/incoming'; $uploadfile = $uploaddir . basename($_FILES['userfile']['name']); echo '<pre>'; if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) { echo "File is valid, and was successfully uploaded.\n"; } else { echo "Possible file upload attack!\n"; } echo 'Here is some more debugging info:'; print_r($_FILES); print "</pre>"; ?> [/code] the problem is,, once I choose the file and click the 'send this file' button, it works, seems fine, says it is succesful, however the file is not located in the ftp address. I can access the same address in IE and upload files to it just draging and dropping..does anyone have any idea why this is doing happening? thanx in advanced