我正在尝试使用以下示例的RestTemplate解析RESTFull调用的结果http://thekspace.com/home/component/content/article/57-restful-clients-in-spring-3.html
XML响应是这样的:
<brands>
<brand>
<nodeRef>1111111</nodeRef>
<name>Test</name>
</brand>
</brands>
首先,我这样配置application-context.xml:
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd">
<bean id="restTemplate" class="org.springframework.web.client.RestTemplate">
<property name="messageConverters">
<list>
<bean id="messageConverter" class="org.springframework.http.converter.xml.MarshallingHttpMessageConverter">
<property name="marshaller" ref="xstreamMarshaller" />
<property name="unmarshaller" ref="xstreamMarshaller" />
</bean>
</list>
</property>
</bean>
<bean id="xstreamMarshaller" class="org.springframework.oxm.xstream.XStreamMarshaller">
<property name="aliases">
<props>
<prop key="brand">com.kipcast.dataModel.drugs.bean.BrandViewList</prop>
</props>
</property>
</bean>
</beans>
类com.kipcast.dataModel.drugs.bean.BrandViewList是一个bean,定义了@XStreamAlias("brand")。
下面是我如何做剩下的调用:ApplicationContext applicationContext = new ClassPathXmlApplicationContext("application-context.xml", WebscriptCaller.class);
RestTemplate restTemplate = applicationContext.getBean("restTemplate", RestTemplate.class);
String url = "http://localhost:8081/alfresco/service/search/brand.xml?q={keyword}&alf_ticket={ticket}";
List<BrandViewList> results = (List<BrandViewList>) restTemplate.getForObject(url, List.class, params);
WebscriptCaller.class是我执行这些指令的类。
当我尝试执行它时,getForObject()失败了,我得到了那个异常:
XStream unmarshalling exception; nested exception is com.thoughtworks.xstream.mapper.CannotResolveClassException: brands
我的问题是,我怎么能解决这个问题?为什么会出现这种异常?我如何告诉他跳过根标签?
-------------- 更新 --------------
修复了一些问题,特别是:
List<Brand> brandViewList = (List<Brand>) restTemplate.getForObject(url, Brand.class, params);
但是现在的结果是:
org.springframework.http.converter.HttpMessageNotReadableException: Could not read [class com.kipcast.dataModel.drugs.bean.Brand]; nested exception is org.springframework.oxm.UnmarshallingFailureException: XStream unmarshalling exception; nested exception is com.thoughtworks.xstream.converters.ConversionException: nodeRef : nodeRef
---- Debugging information ----
message : nodeRef
cause-exception : com.thoughtworks.xstream.mapper.CannotResolveClassException
cause-message : nodeRef
class : java.util.ArrayList
required-type : java.util.ArrayList
converter-type : com.thoughtworks.xstream.converters.collections.CollectionConverter
path : /brands/brand/nodeRef
line number : 3
class[1] : com.kipcast.dataModel.drugs.bean.Brands
converter-type[1] : com.thoughtworks.xstream.converters.reflection.ReflectionConverter
version : null
-------------------------------
编辑:更新为只包含相关信息
如果你有不同的类处理"brands"one_answers"brand"标签是最好的。我将创建一个Brand
类,将BrandList
重命名为Brands
(更接近它们引用的XML部分),并让Brands
保存List<Brand>
。给这两个类加上适当的注解,你就可以完成了,例如:
@XStreamAlias("brands")
class Brands {
@XStreamImplicit
List<Brand> brand;
}
@XStreamAlias("brand")
class Brand {
String nodeRef;
String name;
}
上面的代码在将对象编组为XML时工作得很好,但是在将XML反编组为对象时就失败了。为了使其正常工作,您需要告诉编组程序您有哪些带注释的类:
<bean name="marshaller" class="org.springframework.oxm.xstream.XStreamMarshaller">
<property name="autodetectAnnotations" value="true"/>
<property name="annotatedClasses">
<array>
<value>com.kipcast.dataModel.drugs.bean.BrandViewList</value>
<value>com.kipcast.dataModel.drugs.bean.BrandView</value>
</array>
</property>
</bean>
我创建了一个示例项目,用于验证设置
我通过使用ArrayList类型解决了这个问题。因此不需要使用伪类来处理列表。它为我工作了这样的东西(没有使用任何注释):
<bean id="xstreamMarshaller" class="org.springframework.oxm.xstream.XStreamMarshaller">
<property name="aliases">
<props>
<prop key="brands">java.util.ArrayList</prop>
<prop key="brand">com.kipcast.dataModel.drugs.bean.BrandView</prop>
</props>
</property>
</bean>