Question

I have this very simple login form with user and pass and I need to check that both are in a database and I’m using ajax for this:

$("#clientes").bind("submit", function() {
    $(".error_usu").hide();
    $.ajax({
        type        : "POST",
        cache       : false,
        url         : "login.php",
        data        : $(this).serializeArray(),
        success: function(data) {
            switch(data){
                case '1'://exito
                    alert("alert!");
                    window.location.href = "clientes.php";
                break;
                case '2'://pass
                    $(".error_pass").show();
                break;
            }
        }
    });

    return false;
});     

inside login.php I have this:

 <? session_start();
 if(!empty($_POST['usuario']) && !empty($_POST['password']))
 {
include "funciones.php";
$username = $_POST['usuario'];
    $password = $_POST['password'];
$con=conectar();
$checklogin = mysql_query("SELECT * FROM clientes WHERE user = '".$username."' AND pass = '".$password."'",$con) or die (mysql_error());

if(mysql_num_rows($checklogin) == 1){
    $_SESSION['usuario'] = $_POST['usuario'];
    $_SESSION['password'] = $_POST['password'];
    echo '1';
}else{
    echo '2';
}

 }else{
     echo '2';
 }?>

When I check with firebug I see I’m getting a 1 or a 2 but de success function is never working! I tried this before I had the database just checking this way inside login.php and it worked:

if($_POST['usuario']=="fulano" && $_POST['password']=="fulano"){
    echo '1';
}

Is it not working because of the $_SESSION variables? I don’t fully understand how ajax function works.

Answer 1

If it’s not entering the switch statement you could multiply the response with 1, then JavaScript makes it a number instead of text.

success: function(data) {
        data *= 1;
        switch(data){

or

success: function(data) {
        data = parseInt(data);
        switch(data){