JSON 使用 Partial 对象序列化和反序列化 C#答案

【问题标题】：JSON serialize and deserialize C# with Partial objectsJSON 使用 Partial 对象序列化和反序列化 C#
【发布时间】：2020-02-25 17:57:13
【问题描述】：

这是我要寻找的最终 JSON 对象 -

{
  "firstName": "John",
  "lastName": "Smith",
  "age": 27,
  "address": {
    "streetAddress": "21 2nd Street",      
    "city": "New York",
    "state": "NY",
    "postalCode": "10021-3100"         <-- Added by my application
  }
}

在这个例子中，我没有名字、姓氏和地址的一部分。我的逻辑部分只需要注入年龄和地址。有没有办法不复制我这边的整个对象？我可以有这样的对象吗 -

public class AdditionalAddressInfo
{
    public string postalCode { get; set; }
}

public class AdditionalUserInfo
{
    public int age { get; set; }
}

有没有办法将它序列化并添加到传入的 JSON 中。

【问题讨论】：

您是否考虑过为此使用json patch？ There 是它的 .NET 实现，how to article 是 .NET Core。还有第三者package

标签： c# json serialization json.net

【解决方案1】：

您可以使用DynamicObject进行部分反序列化，这样您就不需要了解其他属性：

var jsonString = @"{
                      ""firstName"": ""John"",
                      ""lastName"": ""Smith"",
                      ""address"": {
                        ""streetAddress"": ""21 2nd Street"",      
                        ""city"": ""New York"",
                        ""state"": ""NY"",     
                      }
                    }";

dynamic dynoObject = JsonConvert.DeserializeObject<dynamic>(jsonString);

//adding age
AdditionalUserInfo additionalUserInfo = new AdditionalUserInfo();
additionalUserInfo.age = 27;
dynoObject.age = additionalUserInfo.age;

//adding postalCode
AdditionalAddressInfo additionalAddressInfo = new AdditionalAddressInfo();
additionalAddressInfo.postalCode = "10021 - 3100";
dynoObject.address.postalCode = additionalAddressInfo.postalCode;

var newJson = JsonConvert.SerializeObject(dynoObject);

那么新的Json就会如预期的那样：

{
  "firstName": "John",
  "lastName": "Smith",
  "age": 27,
  "address": {
    "streetAddress": "21 2nd Street",      
    "city": "New York",
    "state": "NY",
    "postalCode": "10021-3100"       
  }
}

【讨论】：

【解决方案2】：

您可以通过修改 JObject 直接将新值添加到 Json。例如，

var jObj = JObject.Parse(json);
var address = jObj["address"] as JObject;
address.Add("postalCode","10021-3100");
jObj.Add("age",27);
var result = jObj.ToString();

输出

{
  "firstName": "John",
  "lastName": "Smith",
  "address": {
    "streetAddress": "21 2nd Street",
    "city": "New York",
    "state": "NY",
    "postalCode": "10021-3100"
  },
  "age": 27
}

【讨论】：