jQuery ajax在前台如何处理后台返回的json数据

Question

jquery ajaxjsonjqueryasp.net

Answer 1

var json = "{\"ok\":true,
                        "ds\":[{\"OrderNo\":\"asdfsdf\",
                        "EmployeeWw\":\"sadfdsf\",
                        "Data\":\"2014-6-3\",
                        "CustomerWw\":\"sdfsadf\",
                        "OrderAttribute\":\"啊啊啊啊啊啊啊啊啊啊啊啊啊啊\",
                        "OrderCount\":\"3\",
                        "OrderMoney\":\"235.00\",
                        "buyertype\":\"True\",
                        "OrderStatus\":\"False\"}]}"json = $.parseJSON(json);
alert(json.ds[0].OrderNo)

Answer 2

var obj = JSON.parse(data);
alert(obj.name);

Answer 3

//success 内方法如下：//判断是否有数据，没有返回
if(result.ok==false) return;
//获取数据
var  rows=result.ds;
for(i=0;i<rows.length;i++)
{
   //要赋什么值你自己选，这里写的是EmployeeWw；
    $("input")[i].val(rows[i].EmployeeWw);
}

Answer 4

 $.ajax({
                    type: "POST",
                    url: "/sales/changeOrder.aspx/GetOrderByID",
                    data: "{id:'" + id + "'}",
                    contentType: "application/json;charset=utf-8",
                    dataType: "json",
                    success: function (result) {
                        //alert(result.d);
                        var obj = JSON.parse(result);
                        alert(obj.ds.OrderNo);
                    },
                    error: function (XMLHttpRequest, data, textStatus) {
                        alert("status:" + XMLHttpRequest.status);
                        alert("readyState:" + XMLHttpRequest.readyState);
                        alert("textStatus:" + textStatus);
                    }
                });

Answer 5

nice ！！！ 太谢谢了  就是这样。