Struts2的初级问题

Posted 2020-10-05 花娣丫头小愤青

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一直报这个错误。终于解决了。

 1 <?xml version="1.0" encoding="UTF-8"?>
 2 <web-app id="WebApp_9" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"
 3          xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
 4          xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
 5 
 6     <filter>
 7         <filter-name>struts2</filter-name>
 8         <filter-class>org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter</filter-class>
 9   <!--    <init-param>  
10        <param-name>filterConfig</param-name>  
11         <param-value>classpath:struts.xml</param-value>  
12    </init-param>  
13   --> 
14     </filter>
15 
16     <filter-mapping>
17         <filter-name>struts2</filter-name>
18         <url-pattern>/*</url-pattern>
19     </filter-mapping>
20 
21     <welcome-file-list>
22         <welcome-file>index.html</welcome-file>
23     </welcome-file-list>
24 
25 </web-app>

web.xml

 1 <?xml version="1.0" encoding="UTF-8" ?>
 2 <!DOCTYPE struts PUBLIC
 3     "-//Apache Software Foundation//DTD Struts Configuration 2.3//EN"
 4     "http://struts.apache.org/dtds/struts-2.3.dtd">
 5 
 6 <struts>
 7     <!-- Struts2 的 Action都必须配置在package里 -->
 8     <package name="default" namespace="/" extends="struts-default">
 9         <action name="loginAction" class="org.crazyit.action.LoginAction">
10         <result name="success">/detils.jsp</result>
11         </action>
12         
13 
14     </package>
15 
16 
17 
18     
19 </struts>

struts.xml

package org.crazyit.action;

public class LoginAction {
    // 封装用户请求参数的username属性
    private String username;
    // 封装用户请求参数的password属性
    private String password;
    public String getUsername() {
        return username;
    }
    public void setUsername(String username) {
        this.username = username;
    }
    public String getPassword() {
        return password;
    }
    
    public void setPassword(String password) {
        this.password = password;
    }
    //处理用户请求的execute方法
    public String execute() throws Exception{
        if(getUsername().equals("wangning")&& getPassword().equals("123456")){
            return "success";
        }else{
            return "error";
        }    
    }
}