Question

我有两个这样的课程：

DataTemplate

我希望得到像这样的JSON：

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;
    }     
}

所以我写了这样的自定义{ "bProp": "bProp", // just serizlised bProp "sProp1": "sProp1_aProp", // computed using aProp "sProp2": "sProp2_aProp" // computed another way }：

JsonSerializer

但我一直收到错误：

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());
    }
}

除非我将com.fasterxml.jackson.core.JsonGenerationException: Can not write a field name, expecting a value和json.writeStartObject();放在serialize方法中（因此它会产生错误的JSON）。

所以我正在寻找像json.writeEndObject();这样的解决方案来与自定义@JsonUnwrapped一起使用。

Answer 1

我理解你的问题，你需要的是UnwrappingBeanSerializer。你可以看到另一个相关的SO帖子： Different JSON output when using custom json serializer in Spring Data Rest

问题在于，您不能在一个字段中同时使用注释@JacksonUnwrapped和@JsonSerialize，因为当您拥有@JsonSerializer时，杰克逊将始终编写字段名称。

以下是完整的解决方案：

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重新定义对象映射器，或者使用其他方法进行研究。

@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组：

public class B {

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


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

Answer 2

我想将此帖子和解决方案添加到此处提出的问题：Using custom Serializers with JsonUnwrapperd因为原始海报正在使用JsonSerializer。 UnwrappingBeanSerializer的建议方法在这种情况下不会起作用。我的帖子目标略有不同，但帖子中的想法应该很容易适用于您的用例，因为它只是覆盖了一个方法而不必在属性上添加除JsonUnwrapped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()));
    }   
}

使用自定义JsonSerializer的Jackson @JsonUnwrapped行为

2 个答案: