使用自定义JsonSerializer的Jackson @JsonUnwrapped行为

Question

我有两个这样的课程:

public class A {
    String aProp = "aProp";

    public String getAProp() {
        return aProp;
    }
}

public class B {
    String bProp = "bProp";
    A a = new A();

    @JsonProperty("bProp")
    public String getBProp() {
        return bProp;
    }

    @JsonSerialize(using = CustomSerializer.class)
    public A getA() {
        return a;
    }     
}

我期待得到像这样的JSON:

{
    "bProp": "bProp",         // just serizlised bProp
    "sProp1": "sProp1_aProp", // computed using aProp
    "sProp2": "sProp2_aProp"  // computed another way
}

所以我写了这样的习惯JsonSerializer:

import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.JsonSerializer;
import com.fasterxml.jackson.databind.SerializerProvider;

public class CustomSerializer extends JsonSerializer<A> {
    @Override
    public void serialize(A a, JsonGenerator json, SerializerProvider provider) throws IOException {
        json.writeStringField("sProp1", "sProp1_" + a.getAProp());
        json.writeStringField("sProp2", "sProp2_" + a.getAProp());
    }
}

但我一直收到一个错误:

com.fasterxml.jackson.core.JsonGenerationException: Can not write a field name, expecting a value

除非我放入json.writeStartObject();并json.writeEndObject();使用序列化方法(因此它产生错误的JSON).

所以我正在寻找一种@JsonUnwrapped与自定义一起使用的解决方案JsonSerializer.

Answer 1

我理解你的问题,你需要的是UnwrappingBeanSerializer.您可以看到另一个相关的SO帖子: 在Spring Data Rest中使用自定义json序列化程序时的不同JSON输出

问题是你不能同时拥有注释@JacksonUnwrapped和@JsonSerialize一个领域,因为当你拥有@JsonSerializer杰克逊时,他总是会写出字段名称.

这是完整的解决方案:

public class CustomSerializer  extends UnwrappingBeanSerializer {
    public CustomSerializer(BeanSerializerBase src, NameTransformer transformer) {
        super(src, transformer);
    }

    @Override
    public JsonSerializer<Object> unwrappingSerializer(NameTransformer transformer) {
        return new CustomSerializer(this, transformer);
    }

    @Override
    protected void serializeFields(Object bean, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonGenerationException {
        A a = (A) bean;
        jgen.writeStringField("custom", a.getAProp());
        jgen.writeStringField("custom3", a.getAProp());
    }

    @Override
    public boolean isUnwrappingSerializer() {
        return true;
    }

}

测试用例,您应该使用自定义配置或其他方法的研究重新定义对象映射器.

@RunWith(SpringJUnit4ClassRunner.class)
@WebAppConfiguration
@SpringApplicationConfiguration(classes = Application.class)
public class ColorsTest {

    ObjectMapper mapper = new ObjectMapper();

    @Before
    public void setUp(){
        mapper.registerModule(new Module() {
            @Override
            public String getModuleName() {
                return "my.module";
            }

            @Override
            public Version version() {
                return Version.unknownVersion();
            }

            @Override
            public void setupModule(SetupContext context) {

                context.addBeanSerializerModifier(new BeanSerializerModifier() {
                    @Override
                    public JsonSerializer<?> modifySerializer(SerializationConfig config, BeanDescription beanDesc, JsonSerializer<?> serializer) {
                        if(beanDesc.getBeanClass().equals(A.class)) {
                            return new CustomSerializer((BeanSerializerBase) serializer, NameTransformer.NOP);
                        }
                        return serializer;
                    }
                });

            }
        });
    }
    @Test
    public void testSerializer() throws JsonProcessingException {
        System.out.println(mapper.writeValueAsString(new B()));
    }
}

B级:

public class B {

        @JsonProperty("bProp")
        public String getBProp() {
            return "bProp";
        }


    @JsonUnwrapped
        public A getA() {
            return new A();
        }
}

Answer 2

我喜欢将这篇文章和解决方案添加到这里提出的问题：使用自定义序列化程序和 JsonUnwrapperd作为原始海报正在使用JsonSerializer。UnwrappingBeanSerializer在这种情况下，使用 的建议方法不起作用。我的帖子有一个稍微不同的目标，但帖子中的想法应该很容易适用于您的用例，因为它只是覆盖了另一种方法，而不必添加除JsonUnwrapped属性之外的一堆东西。

import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.annotation.JsonUnwrapped;
import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.JsonSerializer;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.databind.SerializerProvider;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;

import java.io.IOException;

public class Test {

    static class A {
        String aProp = "aProp";

        public String getAProp() {
            return aProp;
        }
    }

    static class B {
        String bProp = "bProp";
        A a = new A();

        @JsonProperty("bProp")
        public String getBProp() {
            return bProp;
        }

        @JsonSerialize(using = CustomSerializer.class)
        @JsonUnwrapped
        public A getA() {
            return a;
        }
    }


    static class CustomSerializer extends JsonSerializer<A> {
        @Override
        public boolean isUnwrappingSerializer() {
            return true;
        }

        @Override
        public void serialize(A a, JsonGenerator json, SerializerProvider provider) throws IOException {
            json.writeStringField("sProp1", "sProp1_" + a.getAProp());
            json.writeStringField("sProp2", "sProp2_" + a.getAProp());
        }
    }

    public static void main(String... a) throws Exception {
        final ObjectMapper o = new ObjectMapper();
        o.enable(SerializationFeature.INDENT_OUTPUT);
        System.out.println(o.writeValueAsString(new B()));
    }   
}