我正在尝试在此示例http://thekspace.com/home/component/content/article/57-restful-clients-in-spring-3.html之后使用RestTemplate解析RESTFull调用的结果

XML响应是这样的：

<brands>
    <brand>
        <nodeRef>1111111</nodeRef>
        <name>Test</name>
    </brand>
</brands>

首先,我像这样配置了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>

com.kipcast.dataModel.drugs.bean.BrandViewList类是定义了@XStreamAlias(“ brand”)的bean.

这里我如何做剩下的电话：

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是我从中执行这些指令的类.

当我尝试执行该操作时,getForObject()失败,并得到该异常：

XStream unmarshalling exception; nested exception is com.thoughtworks.xstream.mapper.CannotResolveClassException: brands

我的问题是,我该如何解决？为什么会出现这种异常？我如何告诉他跳过根标记？

– – – – – – – 更新 – – – – – – –
解决了一些问题,尤其是：

List<Brand> brandViewList = (List<Brand>) restTemplate.getForObject(url, Brand.class, params);

但现在的结果是：

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
-------------------------------

解决方法:

编辑：更新为仅包含有关信息

最好有不同的类来处理“品牌”和“品牌”标签.我将创建一个Brand类,将BrandList重命名为Brands(以使其更接近他们引用的XML部分),然后让Brands保持List< Brand>.在两个类上都添加正确的注释,然后应该完成操作,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时可以完美地工作,但是当您描述从XML到对象进行编组时,上述代码会失败.为了使它正常工作,您需要告诉编组器您拥有哪些带注释的类：

<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sample project,用于验证设置.