创建JSON JFORM请求

Question

我想用JSON Form Playground从JSON动态创建html表单，这将允许他们编辑已经存在的请求。我有请求保持在db的某个地方，我有映射请求类。我想将json请求转换为JSON Form请求格式。

防爆 - 持久化的请求

{"message":"My message","author":{"name":"author name","gender":"male","magic":36}} 

映射类

public class DummyRequest 
{ 
    @SerializedName("message") 
    private String message; 

    @SerializedName("author") 
    private Author author; 

    // constructors, getters and setters ommitted 
    public static class Author 
    { 
     @SerializedName("name") 
     private String name; 

     @SerializedName("gender") 
     private Gender gender; 

     @SerializedName("magic") 
     private Integer magic; 
    } 

    public static enum Gender 
    { 
     male, female, alien 
    } 
} 

我创建的上述请求，其被持久化为如下：从上述

public static void main(String[] args) 
    { 
     DummyRequest dummyRequest = new DummyRequest(); 
     dummyRequest.setMessage("My message"); 
     DummyRequest.Author author = new DummyRequest.Author("author name", DummyRequest.Gender.male, 36); 
     dummyRequest.setAuthor(author); 
     String dummyRequestJson = new Gson().toJson(dummyRequest); 
     System.out.println(dummyRequestJson); 
    } 

现在，我想创建以下格式的JSON：

{ 
    "schema": { 
    "message": { 
     "type": "string", 
     "title": "Message", 
     "default": "My message" 
    }, 
    "author": { 
     "type": "object", 
     "title": "Author", 
     "properties": { 
     "name": { 
      "type": "string", 
      "title": "Name" 
     }, 
     "gender": { 
      "type": "string", 
      "title": "Gender", 
      "enum": [ "male", "female", "alien" ] 
     }, 
     "magic": { 
      "type": "integer", 
      "title": "Magic number", 
      "default": 42 
     } 
     }, 
     "default": {"name": "Author name", "gender": "alien", "magic": 36} 
    } 
    } 
} 

这似乎相当复杂和乏味，如果我接近蛮力的方式。有人可以指导我如何继续。我不想在Java中创建任何新的请求类。

Answer 1

https://github.com/google/gson

您可以使用模型内的模型来实现你所需要的。

所有你需要做的就是使用GSON如下

 @SerializedName("schema") 
     private Schema schema; 

而且你的Schema对象会像

 @SerializedName("message") 
     private Message message; 
     @SerializedName("author") 
     private Author author; 
      -- and so on 

使用下面的代码从JSON获取对象序列化数据

Gson gson=new Gson(); 
    Model yourModel=gson.fromJson(<your json object as string>); 

使用以下代码从对象中获取json字符串

Gson gson=new Gson(); 
    String string=gson.toJson(yourObject,YourObject.class);