4

在一个典型的 Spring MVC 项目中，我试图访问从外部 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一部分。但它坏了，我必须修理它。也就是说：我对相关代码不是很熟悉。

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>

解组

  • 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ServiceTO类

    @XmlRootElement(name="service")
public class ServiceTO
{
    // [...]
    @XmlElement
    public String title;

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

  • marshaller/unmarshaller 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path.to.extended.jaxb.JaxbJacksonObjectMapper的是：

        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，@XmlElementWrapper(name="services") @XmlElement(name="service")关于 xml 数据包装，第一个清单的注释对我来说看起来不错。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)

类似这样的相关问题已通过注释修复@XmlElementWrapper(name="services")。但这已经存在。

我将不胜感激任何建议。谢谢你。

——马丁

4

1 回答 1

2

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_

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

必须改为

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

whereServiceTOList.class必须是这样的：

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

    public List<ServiceTO> services;
}
于 2011-08-11T20:11:13.897 回答