我此刻正把头撞在墙上。我做了一些开发,使用CXF和Spring创建了一个WebService。Jetty 8运行得很好,但是当我尝试使用Tomcat 6运行它时,我甚至无法从请求中获得WSDL。详情如下:
我的webservice接口:
@WebService(name="AlertService")
public interface AlertWebService {
@WebMethod
HandleAlertResponse openHandleAlert(@WebParam(name = "request") HandleAlertRequest request) throws ServiceException;
}我的webservice进程:
@WebService(endpointInterface = "c.h.g.p.service.ws.alert.AlertWebService", serviceName = "AlertService")
public class AlertWebServiceImpl implements AlertWebService {
@Override
public HandleAlertResponse openHandleAlert(HandleAlertRequest request) throws ServiceException {
//Do some stuff
}My web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app ...>
<display-name>Prime Service Webapp</display-name>
<listener>
<listener-class> org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<servlet>
<servlet-name>CXFServlet</servlet-name>
<servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>CXFServlet</servlet-name>
<url-pattern>/prime-service-webapp/*</url-pattern>
</servlet-mapping>
</web-app>端点定义
<import resource="classpath:META-INF/cxf/cxf.xml" />
<import resource="classpath:META-INF/cxf/cxf-extension-soap.xml" />
<import resource="classpath:META-INF/cxf/cxf-servlet.xml" />
<bean id="alertWebService" class="c.h.g.p.service.ws.alert.AlertWebServiceImpl" />
<jaxws:endpoint implementor="#alertWebService" address="/AlertService" />我点击的网址给了我404个错误
http://localhost:8090/prime-service-webapp/AlertService?wsdl我已经更改了tomcat配置,以查看端口8090。我只是把我的战争叫做黄金服务-webapp放到了tomcat的webapp目录中。在主服务-webapp目录中添加一个index.html文件似乎也可以解决问题。但我根本无法获得wsdl。
有什么想法-告诉我。是否有什么特定于tomcat 6服务器配置的东西,我需要更改才能使其正常工作?
发布于 2014-02-13 18:06:11
我承认我对jetty一无所知,但是在tomcat中,用你的话来说,你需要使用这个URL:
http://localhost:8090/prime-service-webapp/prime-service-webapp/AlertService?wsdl试一试。原因是tomcat默认将war名称指定为上下文根,并且您已经指示cxf servlet侦听
<url-pattern>/prime-service-webapp/*</url-pattern>这意味着上下文根下的另一个级别(即使它是相同的)
https://stackoverflow.com/questions/21761984
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