Ajax调用php响应faild加载响应

Question

您好我正在开发一个chrome应用程序并在amazon elasticbeanstalk中托管了几个文件，并通过ajax向该php文件发送请求。这是我的 ajax 代码

var dataString = "em=" + email + "&ids=" + ids + "&retrieve";
$.ajax({
    type: "POST",
    url: "http://xxxxx.elasticbeanstalk.com/retrieve_item.php",
    data: dataString,
    crossDomain: true,
    cache: false,
    beforeSend: function() {},
    success: function(data1) {
        var ss = jQuery.parseJSON(data1);
        $.ajax({
            type: "POST",
            url: "http://xxxxx.elasticbeanstalk.com/retrieve_stock.php",
            data: dataString,
            crossDomain: true,
            cache: false,
            beforeSend: function() {},
            success: function(datas) {
                if (ss.length == 0) {
                    var sts = "TICKET ID : " + 1;
                    $(".payticket").text(sts);
                } else {
                    var ss1 = jQuery.parseJSON(
                        datas);
                    for (var s = 0; s < ss.length; s++) {
                        var dt = ss[s].ticket_date;
                        var dt1 = dt.split("-");
                        console.log(dt1);
                        var d_day = parseInt(dt1[2]);
                        if (d_day < 10) {
                            d_day = "0" + d_day;
                        }
                        var d_month = dt1[1];
                        var d_year = dt1[0];
                        var ddate = d_year + "-" +
                            d_month + "-" + d_day;
                        console.log(ddate);
                        var todays_date =
                            currentDate.split("/");
                        var t_year = todays_date[2];
                        var t_mnt = todays_date[1];
                        var t_day = parseInt(
                            todays_date[0]);
                        if (t_day < 10) {
                            t_day = "0" + t_day;
                        }
                        var tdate = t_year + "-" +
                            t_mnt + "-" + t_day;
                        console.log(tdate);
                        var gid = ss[s].id;
                        var gids = ss[0].id;
                        var gg = parseInt(gids) + 1;
                        var gg1 = "TICKET ID : " +
                            gg;
                        $(".payticket").text(gg1);
                        if (ddate == tdate) {
                            var stat = ss[s].ticket_status;
                            var cn = ss[s].client_name;
                            var tot = ss1[s].final_total;
                            var tstat;
                            var cls;
                            if (stat == "open") {
                                tstat = "OPEN";
                                cls = "open";
                            } else {
                                tstat = "CLOSED"
                                cls = "open1";
                            }
                            $("#cashreg1").append(
                                '<tr id="' +
                                gid +
                                '"><td><div class="backshadow"><div class="cells1"><div class="titlebar"><label class="ticket">TICKET ID - ' +
                                gid +
                                '</label><label class=' +
                                cls + '>' +
                                tstat +
                                '</label></div><input type="checkbox" class="check_condition" style="display:none;"><div class="whitecontent"><label class="name">' +
                                cn +
                                '</label><label class="close">' +
                                tot +
                                '</label></div></div></div></td></tr>'
                            );
                        }
                    }
                }
            }
        });
    }
});

首先ajax响应数据，但外部ajax成功的另一个ajax调用无法响应。 ajax有什么问题吗？我删除了dataType =＆＃39; json＆＃39;并尝试添加它，但我没有得到任何回应。 这是我的 php 代码

$em=$_POST['email'];
 $sid=$_POST['ids'];
$data=array();
$q1="SELECT * FROM `".$em."`.`".$em."_".$sid."_single_item`";
$q = mysqli_query($sql,$q1);
while ($row=mysqli_fetch_object($q)){
 $data[]=$row;
}
echo json_encode($data);

我返回json数据。上面的代码有什么问题吗？有人可以帮助我吗？