Question

我的模型如下所示，其中bookJson是一个json对象 -

{
    "name" : "somebook",
    "author" : "someauthor"
}

public class Book{
    private int id;
    private JSONObject bookJson;

   public int getId(){return this.id;}
   public JSONObject getBookJson(){this.bookJson;}
   public void setId(int id){this.id = id;}
   public void setBookJson(JSONObject json){this.bookJson = json;}

}

JSONObject属于org.json包

当My RestController在ResponseEntity对象中返回Book对象时，我收到错误 -

 "errorDesc": "Type definition error: [simple type, class org.json.JSONObject]; nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: No serializer found for class org.json.JSONObject and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS)

实现这一目标的最佳方法是什么？

如果没有一个包含bookJson所有字段的模型类，我们难道不能实现这个目标吗？

Answer 1

想出来了。

将JSONObject添加为Map，并使用ObjectMapper进行所有转换。

public class Book{
    private int id;
    private Map<String, Object>bookJson;

   public int getId(){return this.id;}
   public Map<String, Object>getBookJson(){this.bookJson;}
   public void setId(int id){this.id = id;}
   public void setBookJson(Map<String, Object> json){this.bookJson = json;}

}



 ObjectMapper mapper = new ObjectMapper();
    try {
        map = mapper.readValue(json, new TypeReference<Map<String, Object>>(){});
    } catch (IOException e) {
        throw new JsonParsingException(e.getMessage(), e);
    }

Answer 2

您需要返回List<Book>类型的ResponseEntity。

public ResponseEntity<List<Book>> doSomething() {
  return new ResponseEntity(bookList);
}

使用ResponseEntity返回一个对象，将JSONObject作为对象的字段之一

2 个答案: