ssh后台URL直接访问Action 404错误

Question

struts.xml

```<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE struts PUBLIC
"-//Apache Software Foundation//DTD Struts Configuration 2.3//EN"
"http://struts.apache.org/dtds/struts-2.3.dtd">

<package name="ssh" extends="struts-default" namespace="/">

    <action name="UserAction" class="UserAction" ></action>
</package>

applicationContext

<!-- 配置action -->

<bean id="UserAction" class="cn.crm.ssh.action.UserAction" scope="prototype">
    <property name="UserService" ref="UserService"></property>
</bean>

web.xml


```<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" 
    xmlns="http://java.sun.com/xml/ns/javaee" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 
    http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">


   <!-- spring核心监听器 -->
  <listener>
  <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>

  <context-param>
  <param-name>contextConfigLocation</param-name>
  <param-value>classpath:applicationContext.xml</param-value>
  </context-param>


  <!-- struts核心过滤器的配置 -->
  <filter>
  <filter-name>struts</filter-name>
  <filter-class>org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter</filter-class>
  </filter>     

  <filter-mapping>
  <filter-name>struts</filter-name>
  <url-pattern>/*</url-pattern>
  </filter-mapping>



  <display-name></display-name> 
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
</web-app>

访问路径 :http://localhost:8080/ssh_communicate/ssh/UserAction.action
初学ssh求大神指出错误

Answer 1

http://blog.csdn.net/u010523770/article/details/51492395