如何使用json.NET反序列化动态命名的根节点 [英] How to deserialize a dynamically named root node with json.NET
问题描述
这是json文件的示例:
This is an example of the json file:
{
"John Smith": {
"id": "72389",
"email": "johnsmith@gmail.com",
"books": [
{
"id": "0",
"title": "The Hunger Games",
"rating": "5"
},
{
"id": "1",
"title": "Harry Potter and the Order of the Phoenix",
"rating": "3"
},
],
"magazines": [
{
"id": "2",
"title": "National Geographic",
"rating": "1"
},
{
"id": "3",
"title": "Wired",
"rating": "4"
}
],
}
}
请注意,根节点有一个动态名称(John Smith),并且我需要反序列化的每个json都有一个不同的名称. 此json结构要求具有如下设置的类:
Notice the root node has a dynamic name (John Smith), and every json I need to deserialize will have a different name. This json structure would require to have classes setup as follows:
public class RootObject
{
public JohnSmith { get; set; }
}
public class JohnSmith //oops
{
public string id { get; set; }
public string email { get; set; }
public List<Book> books { get; set; }
public List<Magazine> magazines { get; set; }
}
public class Book
{
public string id { get; set; }
public string title { get; set; }
public string rating { get; set; }
}
public class Magazine
{
public string id { get; set; }
public string title { get; set; }
public string rating { get; set; }
}
我的目标是反序列化绕过/忽略"根对象,最重要的是动态地命名节点.这不是很关键,但是我希望能够获得姓氏并将其设置为Person类的一个属性.
My goal is to deserialize "bypassing/ignoring" root object and most importantly that dynamicaly named node. This is not crucial, but I would like to be able to get the last name and set as a property on the Person class.
public class Person
{
public string id { get; set; }
public string email { get; set; }
public string name { get; set; }
public List<Book> books { get; set; }
public List<Magazine> magazines { get; set; }
}
public class Book
{
public string id { get; set; }
public string title { get; set; }
public string rating { get; set; }
}
public class Magazine
{
public string id { get; set; }
public string title { get; set; }
public string rating { get; set; }
}
这是我现在的操作方式:
Here is how I am doing this now:
var jo = JObject.Parse(json);
var deserializable = jo.First.First.ToString();
string name;
var jp = (JProperty)jo.First;
if (jp != null) name = jp.Name;
var person = JsonConvert.DeserializeObject<Person>(deserializable);
person.name = name;
这可以正常工作,但是我想知道,也许可以通过使用自定义JsonConverter来做得更好?恐怕这对我的自动取款机来说有点过头了,所以我想在这里寻求帮助...
This works OK, but I was wondering, maybe it could be done better by using a custom JsonConverter? I'm afraid this is a little bit over my head atm, so I am asking here for some help...
无论如何,如果有更好的方法可以实现这一目标,请分享.
Anyway, if there is any better way to achieve this, please share.
推荐答案
我将保留解决方案的第一部分(反序列化为JObject
),但不会进行其他序列化.我的代码如下:
I would keep the first part of your solution (deserializing to JObject
), but I wouldn't do another serialization. My code would look like this:
var jo = JObject.Parse(json);
var jp = jo.Properties().First();
var name = jp.Name;
var person = jp.Value.ToObject<Person>();
如果需要自定义转换器,可以使用以下代码.转换器将您的对象转换为Person
的列表,其中每个属性代表另一个Person
.
In case you want a custom converter, you could use the following code. The converter converts your object to a list of Person
s where every property represents another Person
.
class PersonListConverter : JsonConverter
{
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
var list = (PersonList) value;
writer.WriteStartObject();
foreach (var p in list.Persons)
{
writer.WritePropertyName(p.Name);
serializer.Serialize(writer, p);
}
writer.WriteEndObject();
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
var jo = serializer.Deserialize<JObject>(reader);
var result = new PersonList();
result.Persons = new List<Person>();
foreach (var prop in jo.Properties())
{
var p = prop.Value.ToObject<Person>();
// set name from property name
p.Name = prop.Name;
result.Persons.Add(p);
}
return result;
}
public override bool CanConvert(Type objectType)
{
return objectType == typeof(PersonList);
}
}
PersonList
的位置如下所示:
[JsonConverter(typeof(PersonListConverter))]
class PersonList
{
public List<Person> Persons { get; set; }
}
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