如何用jackson传递构造函数的参数？ [英] How to pass constructor's parameters with jackson?

问题描述

我正在尝试使用杰克逊去除对象

i am trying to desearlize an object using Jackson

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)

我知道发生了这种情况，因为这是我的构造函数：

I know that this is happening because this is my constructor:

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，我的构造函数收到HttpResponse参数而我不是通过它，但我不知道该怎么做。我不能用空构造函数过度充电，因为我需要以这种方式接收HttpResponse对象。
当我调用readValue（）方法时，有没有办法传递这个构造函数参数？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

So, my constructor recieve HttpResponse parameter and i am not passing it, but i don't know how to do it. I cant overcharge with an empty constructor because i need that to recieve HttpResponse object at this way. Is there any way to pass this constructor param when i call readValue() method? Or what could be the best option at this case? I appreciate your help. Regards

推荐答案

你可以使用 Jackson值注入功能传递一个对象引用，该对象引用不是输入JSON的一部分作为构造函数参数。下面是一个示例：

You can use the Jackson value injection feature to pass an object reference which is not a part of the input JSON as a constructor parameter. Here is an example:

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}

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