java - jackson : How to ignore the POJO name while converting a POJO to JSON using Jackson?

Question

我正在使用 Jackson 2.10.1 库将我的 Java POJO 转换为 JSON，我得到以下输出，我需要没有 POJO 名称的输出(此处为 MyTestPojo)，我尝试了各种 jackson 注释，如 @JsonIgnoreProperties，但那些主要用于 POJO 中存在的成员，而不是 POJO 类名。

{
    "MyTestPojo": [
        {
            "CreatedBy": "user1",
            "Name": "testABC",
            "UpdatedBy": null,
            "UpdatedDate": null,
            "IsActive": true,
            "Value": "testABC1",
            "CreatedDate": "2017-03-13 15:41:54.0",
            "Description": "testABC"
        },
        {
            "CreatedBy": "user2",
            "Name": "testABC",
            "UpdatedBy": null,
            "UpdatedDate": null,
            "IsActive": false,
            "Value": "testABC2",
            "CreatedDate": "2017-03-13 15:41:54.0",
            "Description": "testABC"
        }
    ]
}

而我需要的是 -

[
        {
            "CreatedBy": "user1",
            "Name": "testABC",
            "UpdatedBy": null,
            "UpdatedDate": null,
            "IsActive": true,
            "Value": "testABC1",
            "CreatedDate": "2019-03-13 15:41:54.0",
            "Description": "testABC"
        },
        {
            "CreatedBy": "user2",
            "Name": "testABC",
            "UpdatedBy": null,
            "UpdatedDate": null,
            "IsActive": false,
            "Value": "testABC2",
            "CreatedDate": "2020-03-10 15:41:54.0",
            "Description": "testABC"
        }
    ]
}

有没有办法用 Jackson 注释来处理这个问题？

我用过的POJO是-

@JacksonXmlRootElement(localName = "ArrayOfTestPojos")
public class GetResponseVO {

    @JsonProperty("MyTestPojo")
    @JacksonXmlProperty(localName = "MyTestPojo")
    @JacksonXmlElementWrapper(useWrapping = false)
    private ArrayList<MyTestPojo> MyTestPojoList;

    public ArrayList<MyTestPojo> getMyTestPojoList() {
        return MyTestPojoList;
    }

    public void setMyTestPojoList(ArrayList<MyTestPojo> MyTestPojoList) {
        this.MyTestPojoList = MyTestPojoList;
    }

// standard getters and setters

}

和

@JacksonXmlRootElement(localName = "MyTestPojo")
public class MyTestPojo {

    @JsonProperty("Name")
    private String name;

    @JsonProperty("Description")
    private String description;

    @JsonProperty("IsActive")
    private int isActive;

    @JsonProperty("Value")
    private String value = null;

    @JsonProperty("CreatedBy")
    private String createdBy;

    @JsonProperty("CreatedDate")
    private String createdDate;

    @JsonProperty("UpdatedBy")
    private String updatedBy;

    @JsonProperty("UpdatedDate")
    private String updatedDate;

// standard getters and setters.    
}
```````````
I am also generating the XML out of this so you can ignore the annotations relevant to XML.

Answer 1

您可以为此目的使用 JsonValue 注释，基本上是“使用此属性的值而不是序列化容器对象”。它也可以用在 getters 上

@JsonValue indicates that results of the annotated "getter" method (which means signature must be that of getters; non-void return type, no args) is to be used as the single value to serialize for the instance. Usually value will be of a simple scalar type (String or Number), but it can be any serializable type (Collection, Map or Bean).

@JsonValue
@JacksonXmlProperty(localName = "MyTestPojo")
@JacksonXmlElementWrapper(useWrapping = false)
private ArrayList<MyTestPojo> MyTestPojoList;

但这是错误的做法，因为它会生成这样的 JSON，这不是合法的 JSON。

{[{"x":"value"}, ...]}

如果你只想改变 JSON 结构(不影响 xml)，你可以使用 MixIn 来达到这个目的。

public interface JsonMixin {

   @JsonValue
   List<MyTestPojo> getMyTestPojoList();   
}

并将其注册到您的对象映射器并从主类中删除@JsonValue。

objectMapper.addMixIn(GetResponseVO.class, JsonMixin.class);