Jquery ajax函数失败，没有任何理由

Question

我有一个基于网络的移动Android应用程序。我用php + jquery + mysql构建它。 ajax代码失败但没有理由。因此它返回'再试一次'。不幸的是我无法在应用程序中调试它。这是代码：

 $('.get-order-button').live('click', function() {

var musteritel = $('#musteritel').val(), musteriad = $('#musteriad').val(),  musteriadres = $('#musteriadres').val(), musterinotu = $('#siparisnotu').val(), odemesekli =$('#odemesekli option:selected').val() ;

if(musteritel != ''){

    $.ajax({
        type        : 'POST',
        url         : '/enfes/temp-order-sent.php', timeout: 10000, 
        cache       : false,
        data        : 'musteritel='+musteritel+'&musteriadres='+musteriadres+'&musterinotu='+musterinotu+'&odemesekli='+odemesekli+'&musteriad='+musteriad,
        dataType    : 'json',
        beforeSend  : function() {
            showDialog('Yükleniyor...')
        },
        whileLoading: function(xhr) {
            if (xhr && xhr.readyState != 4)
                xhr.abort()
        }
    }).always(function() {
        closeDialog()
    }).fail(function() {
        showToastShort('Try again.');
    }).done(function(r) {

        if(r.s==1){

            $.each(r.u, function( index, value ) {
               unsetMyCookie(index);
              });


            $('#detail').remove();
            $('.basket-added-btn span').text('0');
            basketStatus = 0;

            showToastLong(r.m);
            window.location.hash = 'home';
        }else
            showToastShort(r.m);
    });

}

return false
});

这是php代码：

 <?php
 session_start();
 include_once 'class.render.php';
 include_once 'class.order.php';

 $kendim = new render();



 if(isset($_SESSION["id"]) && isset($_POST['musteritel'])) {

$kendiId = $_SESSION["kendiId"];
$uniqueIdentifier = $_SESSION["uniqueIdentifier"];

if(isset($_COOKIE)){

        $order = new order();

        $musteritel = $order->validTel($_POST['musteritel']);
        $musteriad = $order->cleanStr($_POST['musteriad']);
        $musteriadres = $order->cleanStr($_POST['musteriadres']);
        $musterinotu = $order->cleanStr($_POST['musterinotu']);
        $odemesekli = $order->cleanStr($_POST['odemesekli']);

        $return = array('s' => 0, 'm' => 'Siparis gonderilemedi. Bilgilerinizi kontrol ediniz.');

        $ordersent = $order->addOrder($musteritel, $musteriad, $musteriadres, $musterinotu, $odemesekli);

        if($musteritel == "") {
            $return = array('s' => 1, 'm' => 'Telefon numaranızı hatalı girdiniz.');
        } else {
            if($ordersent){
                $return = array('s' => 1, 'm' => 'Siparisiniz bize ulasmistir.', 'u' => $order->basket);
            }else
                $return = array('s' => 0, 'm' => 'Siparis gonderilemedi. Bilgilerinizi kontrol ediniz.');
        }

}else
    $return = array('s' => 0, 'm' => 'Siparis sepetiniz bos.');

 }else
     $return = array('s' => 0, 'm' => 'Telefon numaranızı yazmalisiniz.');

 echo json_encode($return);
 ?>

为什么会失败？