I have a simple form that I am trying to submit by jQuery. This is it:
<form id="turn_conf" action='turn_insert.php' method="POST">
<label for="nombre">Nombre:</label>
<input type="text" id="tu_name" name="tu_name">
<input type="submit" class="botonEnvio" value="Crear">
</form>
And a jQuery function that by the moment does not seem to work, neither the "It seems I'm working" alert is displayed on screen
$('#turn_conf').on('submit', validaDatos)
function validaDatos(e) {
var nombreTurno = $('input[name=tu_name]').val();
$.post({url: $(this).attr('action'),
data: { tu_name:nombreTurno },
success:feedback })
e.preventDefault(); }
function feedback (datos) {
alert("Seems I'm working!")}
No value is being inserted on DDBB. Here you have what is on the insert PHP file "turn_insert.php":
$nombreTurno = $_POST['tu_name'];
$insertar = mysql_query("INSERT INTO turn_conf (tu_id,tu_name,tu_status) VALUES ('','$nombreTurno','1')");
Console shows error: error on console: POST localhost/Gestion/%5Bobject%20Object%5D 404 Not Found
Any suggestion about what is going wrong here? Thank you in advance.