在Struts2应用程序中获取404错误

时间:2014-12-23 11:26:31

标签: java jsp struts2

我是第一次使用Struts2进行简单的应用程序。

这是我的代码,

ActionExample:

package com.demo;

import com.opensymphony.xwork2.Action;


public class ActionExample implements Action
{

    @Override
    public String execute()  {

        return SUCCESS;
    }

}

success.jsp:

<%@ page language="java" contentType="text/html; charset=UTF-8"
    pageEncoding="UTF-8"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Insert title here</title>
</head>
<body>

Success Page!.........

</body>
</html>

struts.xml:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE struts PUBLIC 
"-//Apache Software Foundation//DTD Struts Configuration 2.0//EN" 
"http://struts.apache.org/dtds/struts-2.0.dtd">

<struts>
<package name="default" extends="struts-default">
<action name="run" class="com.demo.ActionExample">
<result name="success">/success.jsp</result>
</action>
</package>
</struts>  

Web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
  <display-name>StrutsFirstApplication</display-name>
  <welcome-file-list>
    <welcome-file>success.jsp</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>

  <filter>
  <filter-name>struts</filter-name>
  <filter-class>org.apache.struts2.dispatcher.FilterDispatcher</filter-class> 
  </filter>
  <filter-mapping>
  <filter-name>struts2</filter-name>
  <url-pattern>/*</url-pattern>
  </filter-mapping>

</web-app>

我正在通过此视频链接https://www.youtube.com/watch?v=ATrmHSfh6Tc

进行操作

即使我在运行此url时遇到404错误:8080 / StrutsFirstApplication / run.action

为什么以及如何解决这个问题?

3 个答案:

答案 0 :(得分:0)

您没有发布index.jsp。它应该是这样的:

<%@ taglib prefix="s" uri="/struts-tags"%>
...
...
<form action="run">
<s:submit/>
</form>

你的struts.xml文件:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE struts PUBLIC
   "-//Apache Software Foundation//DTD Struts Configuration 2.0//EN"
   "http://struts.apache.org/dtds/struts-2.0.dtd">
<struts>
   <package name="default" extends="struts-default">
<action name="run" class="default.ActionExapmle " method="execute">
        <result name="success">/success.jsp</result>
</action>

  </package>
</struts>

现在,我不知道在web.xml文件中使用所有welcome文件是否正确,请尝试使用:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
  <display-name>Struts 2</display-name>
   <welcome-file-list>
      <welcome-file>index.jsp</welcome-file>
   </welcome-file-list>
   <filter>
      <filter-name>struts2</filter-name>
      <filter-class>
         org.apache.struts2.dispatcher.FilterDispatcher
      </filter-class>
   </filter>

   <filter-mapping>
      <filter-name>struts2</filter-name>
      <url-pattern>/*</url-pattern>
   </filter-mapping>
</web-app>

答案 1 :(得分:0)

配置文件中的问题请检查以下内容并匹配:

<!DOCTYPE struts-config PUBLIC
          "-//Apache Software Foundation//DTD Struts Configuration 1.3//EN"
          "http://jakarta.apache.org/struts/dtds/struts-config_1_3.dtd">
<struts-config>
    <action-mappings>
        <<action name="run" class="com.demo.ActionExample">
            <forward name="success" path="/success.jsp"/>
        </action>

    </action-mappings>

</struts-config>

相应地改变。我建议您切换到struts2,因为struts1是EOL。

答案 2 :(得分:-1)

我已经花了一段时间研究struts,但我认为动作的返回应该与xml配置中的字符串相匹配

我的意思是

public String execute()  {

        return SUCCESS; // returns SUCCESS and not success
    }

你的xml有

<action name="run" class="com.demo.ActionExample">
<result name="success">/success.jsp</result>
</action>

使两端的返回字符串相同,我希望这可能有用。