Question

mvc ajax传json对象后台怎么获取里面的值并放入数据库中

如题，代码贴上
前台代码，：
function saveUser() {
var uname = document.getElementById("uname").value;
var pwd = document.getElementById("pwd").value;
var name = document.getElementById("name").value;
var age = document.getElementById("age").value;
var typeId = document.getElementById("typeId").value;
var doctorpass = "{ \"doctor\":";
doctorpass += "{ \"uname\":\"" + uname + "\",\"pwd\":\"" + pwd + "\",\"name\":\"" + name + "\",\"age\":\"" + age+ "\",\"typeId\":\"" + typeId + "\"}}";

document.getElementById("12").value = doctorpass;

Ajax.request('Doctor/Tjson', {
type: "POST",
data: doctorpass,
success: function (xhr) {
alert("传成功");
document.getElementById("12345").value = xhr.responseText;
// document.getElementById("12345").value = "{ \"uname\":\"" + uname + "\",\"pwd\":\"" + pwd + "\"}";
},
failure: function () {
//to do with xhr
alert("传失败");
}
}
);
后台我想把里面的值取出来，放到数据库中：
public string Tjson(object doctorpass)
{
//怎么接受数据并放到数据库中
}

Answer 1

先看后台action方法
public ActionResult GetValue()
{
ChangePasswordModel model1 = new ChangePasswordModel();
model1.OldPassword = "111";
model1.NewPassword = "222";
return this.Json(model1);//返回对象到前台
}
这样返回的model1对象就会传递到result中去，前台可以直接使用result.NewPassword来取值
如下
$.ajax(
{
url: "GetValue", //表示提交给的action
type: "post", //提交方法
datatype: "json",//数据类型
success: function (result) { //返回的结果自动放在resut里面了
alert(result.NewPassword);
}
});

Answer 2

可以用GSON转一下