解析JSON而不声明模型类 [英] Parse JSON without declaring model class

问题描述

我有这样的JSON:

{
"bookings": {
    "group_id": "abc",
    "name": "Study Rooms",
    "url": "My URL",
    "timeslots": [{
            "room_id": "bcd",
            "room_name": "101",
            "booking_label": "Meeting1",
            "booking_start": "2018-11-30T07:00:00-06:00",
            "booking_end": "2018-11-30T07:30:00-06:00",
            "booking_created": "2018-11-28T11:32:32-06:00"
        }, {
            "room_id": "cde",
            "room_name": "102",
            "booking_label": "Meeting2",
            "booking_start": "2018-11-30T07:30:00-06:00",
            "booking_end": "2018-11-30T08:00:00-06:00",
            "booking_created": "2018-11-28T11:32:32-06:00"
        }, //##AND many more like this##
    ]
}
}

如果我尝试这样解析它:

If I try to parse it like this:

var reservations = new { bookings = new { group_id = "", name = "", url="", timeslots = new List<Timeslot>() } };
Newtonsoft.Json.JsonConvert.PopulateObject(jsonResult, reservations);

仅填充时隙元素

但是，如果我声明一个具有groop_id，名称，URL和时隙集合属性的模型类，并进行如下解析:

However, if I declare a model class with properties groop_id, name, url, and timeslots collection, and parse like this:

var reservations = new { bookings = new BookingsModel() };
Newtonsoft.Json.JsonConvert.PopulateObject(jsonResult, reservations);

它工作正常.

问题是为什么，并且可以在不静态声明模型的情况下解析JSON的所有元素.

Question is WHY, and is it possible to parse all elements of JSON without static declaration of the model.

推荐答案

无法填充匿名对象的原因是，在c#中，匿名类型为

The reason that you are not able to populate your anonymous object is that, in c#, anonymous types are immutable.

相反，您可以使用 JsonConvert.DeserializeAnonymousType() 来创建来自现有实例的匿名类型的新实例:

Instead, you may use JsonConvert.DeserializeAnonymousType() which will create a new instance of your anonymous type from your existing instance:

var reservations = Newtonsoft.Json.JsonConvert.DeserializeAnonymousType(jsonResult, 
        new { bookings = 
            new { group_id = default(string), name = default(string), url=default(string), 
                 timeslots = default(List<Timeslot>) } });      

样本小提琴此处.

这篇关于解析JSON而不声明模型类的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！