使用FasterXML Jackson进行奇怪的JSON序列化

Question

我不明白为什么我得到给定类的序列化JSON，如下所示。

这是从WSDL生成的类，因此我无法更改它：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "Lawyer")
public class Lawyer extends Person {

    @XmlElementWrapper(required = true)
    @XmlElement(name = "lawyerOffice", namespace = "http://xxx/addressbook/external/v01/types")
    protected List<LawyerOffice> lawyerOffices;     

    public List<LawyerOffice> getLawyerOffices() {
    if (lawyerOffices == null) {
        lawyerOffices = new ArrayList<LawyerOffice>();
    }
    return lawyerOffices;
    }

    public void setLawyerOffices(List<LawyerOffice> lawyerOffices) {
    this.lawyerOffices = lawyerOffices;
    }

}

当一个类实例使用fasterxml.jackson序列化时，我得到：

{
    "ID": "e0d62504-4dfb-4c92-b70b-0d411e8ed102",
    "lawyerOffice": [
         {
            ...
         }
     ]
}

因此，数组的名称是律师办公室 。 我希望lawyerOffice 秒 。

这是我使用的实现：

    <dependency>
        <groupId>com.fasterxml.jackson.jaxrs</groupId>
        <artifactId>jackson-jaxrs-json-provider</artifactId>
        <version>2.8.6</version>
    </dependency>

    <dependency>
        <groupId>com.fasterxml.jackson.core</groupId>
        <artifactId>jackson-databind</artifactId>
    </dependency>

这是我的ObjectMapper配置（注入CXF中）：

@Provider
@Consumes({ MediaType.APPLICATION_JSON, "text/json" })
@Produces({ MediaType.APPLICATION_JSON, "text/json" })
public class JsonProvider extends JacksonJsonProvider {

    public static ObjectMapper createMapper() {
    ObjectMapper mapper = new ObjectMapper();

    AnnotationIntrospector primary = new DPAJaxbAnnotationIntrospector(mapper.getTypeFactory());
    AnnotationIntrospector secondary = new JacksonAnnotationIntrospector();
    AnnotationIntrospector pair = AnnotationIntrospector.pair(primary, secondary);
    mapper.setAnnotationIntrospector(pair);
    mapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
    mapper.disable(SerializationFeature.WRITE_SINGLE_ELEM_ARRAYS_UNWRAPPED);
    mapper.disable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY);
    mapper.configure(MapperFeature.ACCEPT_CASE_INSENSITIVE_PROPERTIES, true);

    return mapper;

    }

    public JsonProvider() {
    super();

    this.setMapper(createMapper());

    }

}

如何获得“正确的”列表名称？

Answer 1

我找到了解决方案。 我在objectMapper配置中启用了USE_WRAPPER_NAME_AS_PROPERTY_NAME功能选项：

    ObjectMapper mapper = new ObjectMapper();
    AnnotationIntrospector primary = new JaxbAnnotationIntrospector(mapper.getTypeFactory());
    AnnotationIntrospector secondary = new JacksonAnnotationIntrospector();
    AnnotationIntrospector pair = AnnotationIntrospector.pair(primary, secondary);
    mapper.setAnnotationIntrospector(pair);
    mapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
    ...
    mapper.enable(MapperFeature.USE_WRAPPER_NAME_AS_PROPERTY_NAME); // <-----
    ...