php 与ajax 数据处理的疑问

Question

前端如下 那么php后端怎么接收数据

var o = {};
        o.gender = sex;
        o.birthday = birth;
        o.xing = name;
        o.hour = (hour == null || hour == '') ? -1 : hour;
        o.minute = (minute == null || minute == '') ? -1 : minute;
        o.state = state;//出生状态
        o.wuxing = wuxing;
        o.web_no = '';

        var index = layer.load(1, {
          shade: [0.1,'#fff'] //0.1透明度的白色背景
        });
        $.ajax({
            url: "http://qm.fotang.cn/index.php/Home/Index/order",
            type: "Post",
        data: o,
        dataType: "JSON",
        async: true,
        success: function (d) {
            layer.close(index);
            if (d.code == 1) {
                //location.href = "/name/resultzy/" + d.guid + "?track=984245";
                alert(d.guid);
            } else {
                alert(d);
            }
        },
        error: function () {
           layer.close(index);
            alert("操作异常，请联系客服");

Answer 1

$data = file_get_contents("php://input");