login ajax

<?php
error_reporting(1);
$act=$_GET['act'];
if($act =='done')
{
    die('anda berhasil masuk ');
}

if($act=='check')
{
    foreach($_POST as $n=>$v)$$n=$v;
    if($user=='gunawan'  )
    {
        $a[stat]=1;
/* lakukan proses session dll disini */
    }else{
        $a[stat]=0;
    }

    $a[post]=$_POST;

    $json=json_encode($a);

die($json);
}

?>
<script src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
ASUMSIKAN KAMU MELAKUKAN
<input type=button onclick="login('gunawan')" value='login benar' >
<input type=button onclick="login('Herry')" value='login salah' >
<form id='myForm'>
<input type=text name='user' value='' readonly class='myuser' />
<input value='password' name='password' class='pass' />
</form>
<script>
function login(t)
{
    $(".myuser").val(t);
    checkLogin();
}
function checkLogin()
{
    var selectorform = 'form#myForm';
    var datax = $(selectorform).serialize();
    var request = $.ajax({
          url: "?act=check",
          type: "POST",
          data: datax,
          dataType: "json"
    });

    request.success(function(msg) {
      console.log('data sudah terkirim');
        if(msg.stat==0)
        {
            alert('periksa kembali user anda');
            return false;
        }else{
            alert('terima kasih atas loginnya');
            window.location.href ="?act=done";
        }

    });
}
</script>