ASP.NET MVC如何将JSON对象从View传递给Controller作为参数

var jsonObject = {
    "name": "John",
    "age": 30,
    "city": "New York"
};

var jsonString = JSON.stringify(jsonObject);

$("#hiddenField").val(jsonString);

public class PersonModel
{
    public string Name { get; set; }
    public int Age { get; set; }
    public string City { get; set; }
}

[HttpPost]
public ActionResult SavePerson(string jsonString)
{
    // 将JSON字符串转换为PersonModel实例
    var person = JsonConvert.DeserializeObject<PersonModel>(jsonString);

    // 在此处处理PersonModel实例

    return View();
}

$.ajax({
    url: "/ControllerName/SavePerson",
    type: "POST",
    data: { jsonString: $("#hiddenField").val() },
    success: function (result) {
        // 处理成功的响应
    },
    error: function (xhr, status, error) {
        // 处理错误的响应
    }
});