java - 如何使用构建器 Jackson 反序列化包含泛型类型的对象？

Question

我目前正在使用 dropwizard 开发一个小型网络应用程序。另一个应用程序可以向我的应用程序发送多种类型的消息。这些消息具有共同的基础和特定的内容，具体取决于消息的类型。

这是消息的通用基础:

@JsonDeserialize(builder = BaseMessageBuilder.class)
public abstract class BaseMessage<T> {
    private String commonFiled;
    private T content;

    protected BaseMessage(BaseMessageBuilder<T> builder) {
        this.commonFiled = builder.getCommonField();
        this.content = builder.getContent();
    }

    public String getCommonField() {
        return commonFiled;
    }

    public T getContent() {
        return content;
    }
}

这是消息公共(public)基础的构建器:

@JsonPOJOBuilder
public class BaseMessageBuilder<T> {
    private String commonFiled;
    private T content;

    public String getCommonField() {
        return commonFiled;
    }

    public BaseMessageBuilder<T> withCommonField(String commonFiled) {
        this.commonFiled = commonFiled;
        return this;
    }

    public String getContent {
        return participants;
    }

    public BaseMessageBuilder<T> withContent(T content) {
        this.content = content;
        return this;
    }

    public BaseMessage<T> build() {
        return new BaseMessage<T>(this);
    }
}

具体内容如下:

@JsonDeserialize(builder = SpecificContentBuilder.class)
public abstract class SpecificContent{
    private String field1;
    private Long field2;

    protected SpecificContent(SpecificMessageBuilder builder) {
        this.field1 = builder.getField1();
        this.field2 = builder.getField2();
    }

    public String getField1() {
        return field1;
    }

    public Long getField2() {
        return field2;
    }
}

下面是具体内容的构建器:

@JsonPOJOBuilder
public class SpecificContentBuilder {
    private String field1;
    private Long field2;

    public String getField1() {
        return commonFiled;
    }

    public SpecificContentBuilder withField1(String field1) {
        this.field1 = field1;
        return this;
    }

    public String getField2 {
        return field2;
    }

    public SpecificContentBuilder withField2(Long field2) {
        this.field2 = field2;
        return this;
    }

    public BaseMessage build() {
        return new BaseMessage<T>(this);
    }
}

这是我的资源:

@Path("/test")
@Produces(MediaType.APPLICATION_JSON)
@Consumes(MediaType.APPLICATION_JSON)
public class TestResource {

    @POST
    public void test(BaseMessage<SpecificContent> message) {
        // ...
    }

}

目前，当我的资源获取对象时，内容消息实际上并未获取正确的类型。当我调试它时，它自动成为一个 LinkedHashMap。我想有什么不对劲。谁能帮帮我？

编辑:

这是一个 Jackson 问题(查看 pandaadb 响应)。我删除了我的 BaseMessageBuilder 并使用了静态构造函数以保持我的对象不可变。

public abstract class BaseMessage<T> {
    private String commonFiled;
    private T content;

    @JsonCreator
    public static <T> BaseMessage<T> newInstance(
            @JsonProperty("commonFiled") String commonFiled,
            @JsonProperty("content") T content) {
        return new WebhookEventBuilder()
                .withCommonFied(commonFiled)
                .withContent(content)
                .build();
    }

    protected BaseMessage(BaseMessageBuilder<T> builder) {
        this.commonFiled = builder.getCommonField();
        this.content = builder.getContent();
    }

    public String getCommonField() {
        return commonFiled;
    }

    public T getContent() {
        return content;
    }

    private static class BaseMessageBuilder<T> {
        private String commonFiled;
        private T content;

        public WebhookEventType getCommonFiled() {
            return commonFiled;
        }

        public BaseMessageBuilder withCommonFiled(String commonFiled) {
            this.commonFiled = commonFiled;
            return this;
        }

        public T getContent() {
            return content;
        }

        public BaseMessageBuilder withContent(T content) {
            this.content = content;
            return this;
        }

        public BaseMessage build() {
            return new BaseMessage(this);
        }
    }
}

Answer 1

试过了，目前似乎不支持:

https://github.com/FasterXML/jackson-databind/issues/921

但是，对于您的示例，这是不必要的。 Jackson 默认做正确的事情，例如，这个代码示例对我有用:

我有 1 个 BaseMessage:

public class BaseMessage<T> {

    @JsonProperty("val1")
    String val1;
    @JsonProperty("val2")
    T val2;
}

还有 2 种不同的内容类型:

public class Content {

    @JsonProperty("val1")
    String val1;
    @JsonProperty("val2")
    long val2;

}

public class Content2 {

    @JsonProperty("val1")
    String val1;
    @JsonProperty("val2")
    String val2;

}

有了这个资源:

@Path("/builder")
@Produces(MediaType.APPLICATION_JSON)
public class BuilderResource {


    @POST
    @Path("/test")
    @Consumes(MediaType.APPLICATION_JSON)
    @Produces(MediaType.APPLICATION_JSON)
    public Response test(BaseMessage<Content> testContent) {

        System.out.println("hit normal content");
        return Response.ok().build();
    }


    @POST
    @Path("/test2")
    @Consumes(MediaType.APPLICATION_JSON)
    @Produces(MediaType.APPLICATION_JSON)
    public Response test2(BaseMessage<Content2> testContent) {

        System.out.println("hit String content");
        return Response.ok().build();
    }
}

点击测试 url，jackson 成功检测到基本内容的类型 T 并创建正确的实例。

另一种选择是为 BaseMessage 类型编写自己的 MessageBodyReader

希望对你有帮助

阿图尔