Question

我的JSON结构如下：

{
  "id" : "123",
  "name" : [ {
    "stuff" : [ {
      "id" : "234",
      "name" : "Bob"
    }, {
      "id" : "345",
      "name" : "Sally"
    } ]
  } ]
}

我想要映射到以下数据结构：

MyInterface1

@Value.Immutable
@JsonSerialize(as = ImmutableMyInterface1.class)
@JsonDeserialize(as = ImmutableMyInterface1.class)
public interface MyInterface1 {
    String id();
    @JsonDeserialize(using = MyInterface1Deserializer.class)
    List<MyInterface2> name();
}

MyInterface2

@Value.Immutable
@JsonSerialize(as = ImmutableMyInterface2.class)
@JsonDeserialize(as = ImmutableMyInterface2.class)
public interface MyInterface2 {
    @JsonDeserialize(using = StuffDeserializer.class)
    Map<String, MyInterface3> stuff();
}

MyInterface3

@Value.Immutable
@JsonSerialize(as = ImmutableMyInterface3.class)
@JsonDeserialize(as = ImmutableMyInterface3.class)
public interface MyInterface3 {
      String id();
      String name();
}

我使用带有readValue(stringWithJson,MyInterface1.class)的ObjectMapper将此JSON映射到MyInterface1，它应该从链接到MyInterface3。当我在MyInterface2中使用List时，此设置正常工作，即List<MyInterface3> name();

但是，我希望这是一张地图而不是一张清单，理想情况下是＆＃34; id＆＃34;从内部JSON作为关键。这将允许我使用以下语法获取值： MyInterface1.get(0).MyInterface2.get("id1").name();

问题是，在尝试创建自定义StuffDeserializer.class时，我收到错误： Can not deserialize instance of com.foo.ImmutableMyInterface2$Json out of START_ARRAY token

尝试时：

public Map<String, MyInterface3> deserialize(JsonParser jsonParser, DeserializationContext ctxt)
        throws IOException {

    MyInterface2 foo = Unmarshaller.OBJECT_MAPPER.readValue(jsonParser, MyInterface2.class); // error here
    ...

我认为这是因为杰克逊期待＆＃34;东西＆＃34;成为JSON数组的List＆＃39;原因。将此JSON反序列化为使用内部JSON中的值作为键的映射的最佳方法是什么？

Answer 1

我会创建一个自定义JsonDeserializer来将id和name映射到地图中：

public class StringHashMapValueDeserializer extends JsonDeserializer<HashMap<String, String>>{

    @Override
    public HashMap<String, String> deserialize(JsonParser parser, DeserializationContext ctxt)
            throws IOException, JsonProcessingException {
        HashMap<String, String> ret = new HashMap<String, String>();

        ObjectCodec codec = parser.getCodec();
        TreeNode node = codec.readTree(parser);

        if (node.isArray()){
            for (JsonNode n : (ArrayNode)node){
                JsonNode id = n.get("id");
                if (id != null){
                    JsonNode name = n.get("name");
                    ret.put(id.asText(), name.asText());
                }
            }
        }
        return ret;
    }
}

然后我会使用反序列化器创建带有注释stuff属性的简单bean：

@Getter
@Setter
public class Name {

    @JsonDeserialize(using = StringHashMapValueDeserializer.class)
    Map<String, String> stuff;

    @Override
    public String toString() {
        return "Name [stuff=" + stuff + "]";
    }
}

外型：

@Getter
@Setter
public class OuterType {

    String id;
    List<Name> name;

    @Override
    public String toString() {
        return "OuterType [id=" + id + ", stuff=" + name + "]";
    }
}

反序列化：

ObjectMapper mapper = new ObjectMapper();

OuterType response;
response = mapper.readValue(json, OuterType.class);

System.out.println(response);
System.out.println(response.getName().get(0).getStuff().get("234"));

控制台输出：

OuterType [id=123, stuff=[Name [stuff={234=Bob, 345=Sally}]]]
Bob

希望它有所帮助。

使用Jackson将JSON数组反序列化为Map

1 个答案: