spring - "Unrecognized field (..), not marked as ignorable"当 jaxb 解码 xml 输入时

Question

在一个典型的 Spring MVC 项目中，我尝试访问从外部 Web 服务源获取的对象。直到现在，这些数据的实际集成实际上并不是我在项目中的部分。但它坏了，我必须修理它。也就是说:我对相关代码并不完全熟悉。

背景

数据

从外部 Web 服务接收的 XML 数据如下所示:

<offeredServiceTOes>
   <OfferedService deleted="false">
      <id>0001_01-u001/igd</id>
      <title>Umschlagleistung (001)</title>
      <mainType>turnover</mainType>
      <services>
         <service id="tos5yyeivg">
            <title>Umschlag Bahn - Binnenschiff</title>
            <mainType>turnover</mainType>
            <systemId>RailRiver</systemId>
            <meansOfTransport id="motRail">
               <title>Bahn</title>
               <description>Bahn</description>
               <systemId>Rail</systemId>
            </meansOfTransport>
            <meansOfTransportRel id="motRiver">
               <title>Binnenschiff</title>
               <description>Binnenschiff</description>
               <systemId>River</systemId>
            </meansOfTransportRel>
         </service>
         <service id="tos5yyeiw0">
            [...]
         </service>
         [...]
      </services>
      [...]
    </OfferedService>
    [...]
<offeredServiceTOes>

解码

• 使用 Spring Rest 模板的方法如下所示:

  @Override
  public List<OfferedServiceTO> getOfferedServices() {
      return restTemplate.getForObject(
              dataServiceUriTemplate, 
              OfferedServiceTOList.class,
              OFFERED_SERVICES
      );
  

• 相关的OfferedServiceTOList类:

  @XmlRootElement(name="OfferedService")
  public class OfferedServiceTO
  {
  
      @XmlElement
      @XmlID
      public String id;
  
      // [...]
  
      @XmlElementWrapper(name="services")
      @XmlElement(name="service")
      public List<ServiceTO> services; 
  
      // [...]
  }
  

• 相关的ServiceTO类

  @XmlRootElement(name="service")
  public class ServiceTO
  {
      // [...]
      @XmlElement
      public String title;
  
      /[...]
      @XmlElementWrapper(name="mainServices")
      @XmlElement(name="service")
      public List<ServiceTO> mainServices;
  }
  

• 编码器/解码器 xml bean 配置

  <bean id="jaxbMarshaller" class="org.springframework.oxm.jaxb.Jaxb2Marshaller">
      <property name="classesToBeBound">
          <list>
              <value>a.b.beans.ServiceTO</value>
              <value>a.b.OfferedServiceTO</value>
              [...]
          </list>
      </property>
  </bean>
  
  <bean id="xmlMessageConverter"
      class="org.springframework.http.converter.xml.MarshallingHttpMessageConverter">
      <constructor-arg ref="jaxbMarshaller" />
  </bean>
  
  <bean id="jsonHttpMessageConverter"
      class="org.springframework.http.converter.json.MappingJacksonHttpMessageConverter">
      <property name="objectMapper" ref="jaxbJacksonObjectMapper"/>
  </bean>
  
  <bean id="jaxbJacksonObjectMapper"
      class="a.b.path.to.extended.jaxb.JaxbJacksonObjectMapper">
  </bean>
  
  <bean id="jsonView" class="org.springframework.web.servlet.view.json.MappingJacksonJsonView">
      <property name="objectMapper" ref="jaxbJacksonObjectMapper" />
  </bean>
  

• 最后，上面提到的path.to.extended.jaxb.JaxbJacksonObjectMapper是:

      public class JaxbJacksonObjectMapper extends ObjectMapper {
  
          public JaxbJacksonObjectMapper() {
              final AnnotationIntrospector primary = new JaxbAnnotationIntrospector();
              final AnnotationIntrospector secondary = new JacksonAnnotationIntrospector();
              AnnotationIntrospector introspector = new AnnotationIntrospector.Pair(primary, secondary);
              DeserializationConfig deserializationConfig = super.getDeserializationConfig().withAnnotationIntrospector(introspector);
              DeserializationProblemHandler errorHandler = new DeserializationProblemHandler() {
                  @Override
                  public boolean handleUnknownProperty(DeserializationContext ctxt, JsonDeserializer<?> deserializer, Object beanOrClass,
                          String propertyName) throws IOException, JsonProcessingException {
                      //TODO Logging (unbekanntes Input-JSON)
                      ctxt.getParser().skipChildren();
                      return true;
                  }
              };
              deserializationConfig.addHandler(errorHandler );
              super.setDeserializationConfig(deserializationConfig);
              SerializationConfig serializationConfig = super.getSerializationConfig().withAnnotationIntrospector(introspector);
              serializationConfig.set(Feature.WRAP_ROOT_VALUE, true);
              super.setSerializationConfig(serializationConfig);
          }
  
      }
  

问题

问题是，第一个列表的注释 @XmlElementWrapper(name="services") @XmlElement(name="service") 在 xml 数据包装方面对我来说看起来很好。但我不断收到错误:

[...] nested exception is org.springframework.web.client.ResourceAccessException: I/O error: Unrecognized field "service" (Class a.b.ServiceTO), not marked as ignorable
 at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@76d697d9; line: 7, column: 18] (through reference chain: a.b.OfferedServiceTO["services"]->a.b.ServiceTO["service"]); nested exception is org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "service" (Class a.b.ServiceTO), not marked as ignorable
 at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@76d697d9; line: 7, column: 18] (through reference chain: a.b.OfferedServiceTO["services"]->a.b.ServiceTO["service"])
    at org.springframework.beans.factory.support.BeanDefinitionValueResolver.resolveReference(BeanDefinitionValueResolver.java:328)

类似的相关问题如this one通过注释 @XmlElementWrapper(name="services") 修复。但这已经存在了。

如有任何建议，我将不胜感激。谢谢。

-- 马丁

Answer 1

好的，这比预期的要容易。 List 字段需要一个包装层。仔细查看json文档发现了解决方案:

在OfferedServiceTO.class中

@XmlElementWrapper(name="services")
@XmlElement(name="service")
public List<ServiceTO> services;

必须更改为

@XmlElement(name="services")
public ServiceTOList services;

其中 ServiceTOList.class 必须类似于:

@XmlRootElement(name="service")
public class ServiceTOList extends ArrayList<ServiceTO> {

    public List<ServiceTO> services;
}