Question

我想使用任意字段序列化和反序列化对象。我尝试将具有任意字段的对象扩展为Dictionary<string, object>，以期能够将任意字段设置为Dictionary条目。在这种情况下，除了Company和Position字段之外，我希望json响应中（在代码注释中列出）还有manager和office。不幸的是，我可以获取任意字段，但无法获取非任意字段。

我还希望能够添加任意对象，而不仅仅是字符串（即向工作添加基本工资，奖金等的薪水对象）。我也有一些限制，因此不能使用dynamic。

public class Person
{
    public string Name { get; set; }
    public int Age { get; set; }
    public Job Job { get; set; }
}

public class Job : Dictionary<string, object>
{
    public string Company { get; set; }
    public string Position { get; set; }
}

var job = new Job()
{
    Company = "Super Mart",
    Position = "Cashier"
};

// Set arbitrary fields
job["manager"] = "Kathy";
job["office"] = "001";

var john = new Person()
{
    Name = "John Doe",
    Age = 41,
    Job = job
};

var employeeJson = JsonConvert.SerializeObject(john, Formatting.Indented);
Log.Debug("TestSerialization", "json: {0}", employeeJson);
// Result
// {
//   "Name": "John Doe",
//   "Age": 41,
//   "Job": {
//     "manager": "Kathy",
//     "office": "001"
//   }
// }

var johnDoe = JsonConvert.DeserializeObject<Person>(employeeJson);
Log.Debug("TestSerialization", "name: {0}, age: {1}", johnDoe.Name, johnDoe.Age);
// Result
// name: John Doe, age: 41

Log.Debug("TestSerialization", "company: {0}, position: {1}", johnDoe.Job.Company, johnDoe.Job.Position);
// Result
// company: , position:

Log.Debug("TestSerialization", "manager: {0}, office: {1}", johnDoe.Job["manager"], johnDoe.Job["office"]);
// Result
// manager: Kathy, office: 001

我使用此代码进行反序列化的结果json是

{
  "Name": "John Doe",
  "Age": 41,
  "Job": {
    "manager": "Kathy",
    "office": "001"
  }
}

我希望结果json是（服务期望的）

{
  "Name": "John Doe",
  "Age": 41,
  "Job": {
    "Company" = "Super Mart",
    "Position" = "Cashier"
    "manager": "Kathy",
    "office": "001"
  }
}

Answer 1

我认为您的工作类别有问题，它是从词典派生的，因此在序列化时将不考虑其成员。仅字典值，

尝试这种方式，我不确定这是否会帮助您解决问题

public class Job 
{
    public string Company { get; set; }
    public string Position { get; set; }
    public Dictionary<string,object> job { get; set; }

    public Job()
    {
        job = new Dictionary<string, object>();
    }
}

Debug.Log（johnDoe.Job.job [“ manager”] +“-” + johnDoe.Job.job [“ office”]）;

在Unity

1 个答案: