如何用杰克逊传递构造函数的参数？

Question

我正在尝试使用 Jackson 对对象进行去耳化

this.prepareCustomMapper().readValue(response.getBody(), EmailResponse.class);

我有这个例外：

org.codehaus.jackson.map.JsonMappingException: No suitable constructor found for type [simple type, class com.despegar.social.automation.services.emailservice.response.EmailResponse]: can not instantiate from JSON object (need to add/enable type information?)
 at [Source: java.io.StringReader@4f38f663; line: 1, column: 12] (through reference chain: com.despegar.social.automation.services.emailservice.response.EmailsResponse["items"])
    at org.codehaus.jackson.map.JsonMappingException.from(JsonMappingException.java:163)
    at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeFromObjectUsingNonDefault(BeanDeserializer.java:746)
    at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:683)
    at org.codehaus.jackson.map.deser.BeanDeserializer.deserialize(BeanDeserializer.java:580)
    at org.codehaus.jackson.map.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:217)

我知道这是因为这是我的构造函数：

public class EmailResponse extends MyServiceResponse {

    private String id;
    private String user_id;
    private String email;
    private Boolean is_primary;
    private Boolean is_confirmed;

    public EmailResponse(HttpResponse request) {
        super(request);
    }
}

所以，我的构造函数接收到 HttpResponse 参数并且我没有传递它，但我不知道该怎么做。我不能用空的构造函数多收费，因为我需要它以这种方式接收 HttpResponse 对象。 当我调用 readValue() 方法时，有什么方法可以传递这个构造函数参数吗？或者在这种情况下最好的选择是什么？我感谢您的帮助。问候

Answer 1

您可以使用the Jackson value injection feature 将不属于输入 JSON 的对象引用作为构造函数参数传递。这是一个例子：

public class JacksonInjectExample {
    private static final String JSON = "{\"field1\":\"value1\", \"field2\":123}";

    // HttpResponse in your case
    public static class ExternalObject {
        @Override
        public String toString() {
            return "MyExternalObject";
        }
    }

    public static class Bean {
        // make fields public to avoid writing getters in this example
        public String field1;
        public int field2;

        private ExternalObject external;

        public Bean(@JacksonInject final ExternalObject external) {
            this.external = external;
        }

        @Override
        public String toString() {
            return "Bean{" +
                    "field1='" + field1 + '\'' +
                    ", field2=" + field2 +
                    ", external=" + external +
                    '}';
        }
    }

    public static void main(String[] args) throws IOException {
        final ObjectMapper mapper = new ObjectMapper();
        final InjectableValues.Std injectableValues = new InjectableValues.Std();
        injectableValues.addValue(ExternalObject.class, new ExternalObject());
        mapper.setInjectableValues(injectableValues);

        final Bean bean = mapper.readValue(JSON, Bean.class);
        System.out.println(bean);
    }
}

输出：

Bean{field1='value1', field2=123, external=MyExternalObject}

Answer 2

您可以编写自定义反序列化器：http://jackson.codehaus.org/1.5.7/javadoc/org/codehaus/jackson/map/annotate/JsonDeserialize.html

在这种情况下，您将能够将所需的任何值传递给构造函数。您需要在 EmailResponse 上添加 @JsonDeserialize 注释，例如：

@JsonDeserialize(using = EmailResponseDeserializer.class)

反序列化器实现示例：

public class EmailResponseDeserializer extends JsonDeserializer<EmailResponse> {
    HttpResponse httpResponse;

    public EmailResponceDeserializer(HttpResponse httpResponse) {
        this.httpResponse = httpResponse;
    }

    @Override
    public EmailResponse deserialize(JsonParser jp, DeserializationContext ctxt) 
      throws IOException, JsonProcessingException {
        JsonNode node = jp.getCodec().readTree(jp);
        int id = (Integer) ((IntNode) node.get("id")).numberValue();
        String email = node.get("email").asText();

        EmailResponse emailResponse = new EmailResponse(httpResponse)
        emailResponse.setId(id);
        emailResponse.setEmail(email);
        // other properties

        return emailResponse;
    }
}

您还需要注册自定义反序列化器：

ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule();
module.addDeserializer(EmailResponse.class, new EmailResponseDeserializer(httpRespose));
mapper.registerModule(module);

一般来说，我会说通过将 HttpResponse 添加到 EmailRespose bean 中，您正在向 DTO 对象中添加一些不应该有的实现。我认为将 httpResponse 设置为自定义反序列化器然后将其设置为 EmailResponse 不是一个好主意，但没有什么能阻止您这样做。

希望这会有所帮助。