php ajax.. with debug

<?php
error_reporting(1);
if($_GET)
{
    ob_start();
    if($_POST)
    {
        echo "isi dari nilai POST yang dikirim \n";
        print_r($_POST);
    }
    $post=ob_get_contents();

    ob_end_clean();
    $a=array(
    'post'=>$post
    );
    echo json_encode($a);
    die($json);
}
?><script src='js/jquery.1.8.2.js'></script>
<script>
function mulaiAjax()
{

var selectorform = 'form#myForm';
/*
perhatikan bahwa penamaan id adalah myForm
*/
    //form-prod-detail
    console.log('mulaiAjax dijalankan');
    var datax = $(selectorform).serialize();

    var request = $.ajax({
          url: "?act=tes",
          type: "POST",
          data: datax,
          dataType: "json"
    });
    request.success(function(msg) {
        console.log('data sudah terkirim');
        $("#hasil").html(msg.post);
    });
    return false;
}
</script>
<form class='myForm' id='myForm'>
<p>Nama<br/><input class='name' id='myName' type=text name='nama' /></p>
<p>ID<br/><input class='id' id='myId' type=text name='id' /></p>
<button onclick='mulaiAjax();return false;'>click disini</button>
</form>
<div id='hasil'></div>