Tomcat上发布webservices的war工程，访问异常404
Tomcat部署正常、war导出工程正常、Tomcat自带的工程可以正常访问；
问题：
webservices工程访问异常404
解决方案：
因为webservices的wsdl访问设置的端口与tomcat端口冲突；
package nc.xyzq.listener;import java.net.InetAddress;import java.net.UnknownHostException;import javax.servlet.ServletContextEvent;import javax.servlet.ServletContextListener;import javax.servlet.annotation.WebListener;import javax.xml.ws.Endpoint;import nc.xyzq.uuib.service.impl.FrServiceImpl; /** ?* @author gacl ?* 用于发布WebService的监听器 ?*/ //使用Servlet.提供的@WebListener注解将实现了ServletContextListener接口的WebServicePublishListener类标注为一个Listener @WebListener public class WebServicePublishListener implements ServletContextListener { ?????@Override ????public void contextDestroyed(ServletContextEvent sce) { ??????System.gc(); ????} ?????@Override ????public void contextInitialized(ServletContextEvent sce) { ???????//WebService的发布地址 ???????String ip =""; ???????try { ???????????ip = InetAddress.getLocalHost().getHostAddress(); ???????} catch (UnknownHostException e) { ???????????// TODO Auto-generated catch block ???????????e.printStackTrace(); ???????} ???????System.out.println("本机的IP = " + ip); ???????String address = "http://"+ip+":8080/webservices/WebService"; ???????//发布WebService，WebServiceImpl类是WebServie接口的具体实现类 ???????Endpoint.publish(address , new FrServiceImpl()); ???????System.out.println("使用WebServicePublishListener发布webservice成功!"); ????} } 端口8080跟tomcat的端口冲突；将wsdl地址的端口改为8060即可；
Tomcat上发布webservices的war工程，访问异常404
原文地址：http://www.cnblogs.com/lizm166/p/7988622.html