创建JSON JFORM请求

Question

我正在尝试使用JSON Form Playground从JSON中动态创建html表单，这将允许他们编辑已经存在的请求。我将请求持久化在db中的某处，并且我有映射请求类。我想将json请求转换为JSON表单请求格式。

非持久化请求

{"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
    }
}

我创建了上面的请求，它被持久化如下：

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，从上面的内容开始，我想创建一个如下格式的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}
    }
  }
}

如果我使用暴力的方式，这似乎相当复杂和乏味。有人可以指导我如何继续下去吗？我不想在Java中创建任何新的请求类。

Answer 1

https://github.com/google/gson

您可以在模型中使用模型来实现所需的功能。

您所需要做的就是使用gson序列化数据，如下所示

     @SerializedName("schema")
       private Schema schema;

您的Schema对象将如下所示

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

使用以下代码从json获取对象

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

使用下面的代码从对象中获取json字符串

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