我如何序列化JSchema作爲另一個對象的一部分？

Question

我有一個對象誰是一個Json Schema（JSchema）的屬性。

JSchema aSchema; 

object foo = new {propA = "x", schema = aSchema}; 

然而，當這是連載：

string str = JsonConvert.SerializeObject(foo); 

的JSchema對象連同其所有其他屬性...而不是一個乾淨的JSON模式序列化，像它的ToString（）的輸出它只是發出Json Schema字符串。

我要的是序列化爲JSON模式的物體，像這樣的架構屬性：

{ 
    "propA": "x", 
    "schema": { 
     "id": "", 
     "description": "", 
     "definitions": { 
      "number": { 
       "type": "number" 
      }, 
      "string": { 
       "type": "string" 
      } 
     }, 
     "properties": { 
      "title": { 
       "title": "Title", 
       "type": "string" 
      } 
     } 
    } 
} 

你會怎麼做呢？

Answer 1

public class Number 
{ 
    public string type { get; set; } 
} 
public class String 
{ 
    public string type { get; set; } 
} 
public class Definitions 
{ 
    public Number number { get; set; } 
    public String @string { get; set; } 
} 
public class Title 
{ 
    public string title { get; set; } 
    public string type { get; set; } 
} 
public class Properties 
{ 
    public Title title { get; set; } 
} 
public class Schema 
{ 
    public string id { get; set; } 
    public string description { get; set; } 
    public Definitions definitions { get; set; } 
    public Properties properties { get; set; } 
} 
public class RootObject 
{ 
    public string propA { get; set; } 
    public Schema schema { get; set; } 
} 

Serialize方法

dynamic coll = new 
{ 
    Root = new RootObject() 
    { 
     propA = "x",    
     schema = new Schema 
     { 
      id = "", 
      description = "", 
      definitions = new Definitions() 
      { 
       number = new Number() 
       { 
        type = "number" 
       }, 
       @string = new String() 
       { 
        type = "number" 
       } 
      }, 
      properties = new Properties() 
      { 
       title = new Title() 
       { 
        title = "Title", 
        type = "string" 
       } 
      } 
     } 
    } 
}; 

var output = JsonConvert.SerializeObject(coll); 

的Fiddler：https://dotnetfiddle.net/f2wG2G

更新

var jsonSchemaGenerator = new JsonSchemaGenerator(); 
var myType = typeof(RootObject); 

var schema = jsonSchemaGenerator.Generate(myType); 
schema.Id = ""; 
schema.Description = ""; 
schema.Title = myType.Name; 

的Fiddler：https://dotnetfiddle.net/FwJX69