我目前正在使用jaxb实现spring web服务。但是当我尝试使用Web服务时,遇到了WebServiceTransportException: Not Found [404]
错误。我确实尝试搜索网络,但无法找到可能的根本原因。下面我展示了我的源代码。
应用程序的context.xml
<bean
class="org.springframework.ws.server.endpoint.adapter.GenericMarshallingMethodEndpointAdapter">
<constructor-arg ref="marshaller" />
</bean>
<bean id="marshaller" class="org.springframework.oxm.jaxb.Jaxb2Marshaller">
<property name="classesToBeBound">
<list>
<value>com.ph.domain.EightBallRequest</value>
<value>com.ph.domain.EightBallResponse</value>
</list>
</property>
</bean>
<bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix">
<value>/jsp/</value>
</property>
<property name="suffix">
<value>.jsp</value>
</property>
</bean>
<bean id="simpleUrlHandlerMapping"
class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping"
lazy-init="true">
<property name="mappings">
<props>
<prop key="/test.asp">LandingController</prop>
</props>
</property>
</bean>
<bean name="LandingController" class="com.ph.controller.LandingController">
<property name="stub" ref="eightBallClient"/>
</bean>
webservice客户端
public class EightBallClient extends WebServiceGatewaySupport {
private Resource request;
public void setRequest(Resource request) {
this.request = request;
}
public String AskQuestion(String question) throws IOException {
String responseString = null;
EightBallRequest request = new EightBallRequest();
request.setQuestion(question);
EightBallResponse response = new EightBallResponse();
response = (EightBallResponse) getWebServiceTemplate()
.marshalSendAndReceive(request);
responseString = response.getAnswer().toString();
return responseString;
}
}
我的网络服务的定义
<bean id="schema" class="org.springframework.xml.xsd.SimpleXsdSchema">
<property name="xsd" value="/WEB-INF/eightball.xsd" />
</bean>
以下是错误堆栈:
SEVERE: Servlet.service() for servlet dispatcher threw exception
org.springframework.ws.client.WebServiceTransportException: Not Found [404]
at org.springframework.ws.client.core.WebServiceTemplate.handleError(WebServiceTemplate.java:626)
at org.springframework.ws.client.core.WebServiceTemplate.doSendAndReceive(WebServiceTemplate.java:550)
at org.springframework.ws.client.core.WebServiceTemplate.sendAndReceive(WebServiceTemplate.java:501)
at org.springframework.ws.client.core.WebServiceTemplate.marshalSendAndReceive(WebServiceTemplate.java:350)
at org.springframework.ws.client.core.WebServiceTemplate.marshalSendAndReceive(WebServiceTemplate.java:344)
at org.springframework.ws.client.core.WebServiceTemplate.marshalSendAndReceive(WebServiceTemplate.java:336)
答案 0 :(得分:1)
也许你的URI:
<bean name="webserviceTemplate" class="org.springframework.ws.client.core.WebServiceTemplate"> <property name="defaultUri" value="http://localhost:8080/mywebservice" />
检查此值:
“HTTP:// mylocal:8080 /为MyWebService”
答案 1 :(得分:1)
以下是我解决此错误的方法:
在marshalSendAndReceive()中使用此回调,如下所示。
final EightBallResponse response = new EightBallResponse();
final SoapActionCallback soapActionCallback = new SoapActionCallback("<the operation name as defined in the WSDL>");
response = (EightBallResponse) getWebServiceTemplate()
.marshalSendAndReceive(request, soapActionCallback );
responseString = response.getAnswer().toString();
答案 2 :(得分:1)
在我的情况下,解决方案是注意URI中的情况。我用小写字母表示,但是web服务期待CamelCase操作名称。