Código PHP:
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if(Upload_Ajax($_FILES['images'],'../../Prueba/'.$Campo['Titulo'].'/')){ $Mysql->Consulta("INSERT INTO Fotos (Foto,Titulo) VALUES ('".$name."','".$Album."')"); } function Upload_Ajax($Files,$Locacion){ foreach($Files['error'] as $key => $error){ if($error == UPLOAD_ERR_OK){ $name = $Files['name'][$key]; return true; } } } }
Metira un error de Notice: undefined variable: name in .....