使用手动配置的方式开发第一个Struts项目的步骤以及实例

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一 先创建一个web工程


 
 
二 引入struts的开发包到项目
http://struts.apache.org下载
 
三 编写login.jsp
<%@ page language="java" import="java.util.*" pageEncoding="ISO-8859-1"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
  <head>
    <base href="<%=basePath%>">
   
    <title>My JSP 'login.jsp' starting page</title>
   
        <meta http-equiv="pragma" content="no-cache">
        <meta http-equiv="cache-control" content="no-cache">
        <meta http-equiv="expires" content="0">   
        <meta http-equiv="keywords" content="keyword1,keyword2,keyword3">
        <meta http-equiv="description" content="This is my page">
        <!--
        <link rel="stylesheet" type="text/css" href="styles.css">
        -->
 
  </head>
 
  <body>
    <form action="/strutslogin/login.do" method="post">
        u:<input type="text" name="username"><br/>
        p:<input type="password" name="password"><br/>
        <input type="submit" value="login">
    </form>
  </body>
</html>
 
四 编写ActonForm(用户表单)和Action(登录小队长)
package com.cakin.forms;
//这是一个用户表单,用于填充数据的
import org.apache.struts.action.ActionForm;
 
public class UserForm extends ActionForm {
        //定义属性【属性名应该和jsp页面的控件名称一致】
        //如果有人提出疑问:表单名字是不是一定要和控件名称一样?不一定,但要保证get和set方法的名字和
        //控件名称一致
        private String username;
        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;
        }
        
}


 
package com.cakin.actions;
//这是一个action(表示小队长,需要继承Action)
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
 
import org.apache.struts.action.Action;
import org.apache.struts.action.ActionForm;
import org.apache.struts.action.ActionForward;
import org.apache.struts.action.ActionMapping;
 
import com.cakin.forms.UserForm;
 
public class LoginAction extends Action {
        //我们需要重新编写一个方法:execute会被自动调用,有点类似servlet的service方法。
        @Override
        public ActionForward execute(ActionMapping mapping, ActionForm form,
                        HttpServletRequest request, HttpServletResponse response)
                        throws Exception {
                        //把form转成对应的UserForm对象
                UserForm userForm=(UserForm)form;
 
                //简单验证
                if("123".equals(userForm.getPassword())){
                        return mapping.findForward("ok");
                }
                else{
                        return mapping.findForward("err");
                }
        }
}
 
五 编写struts-config.xml文件,该文件用于配置action actionForm以及对应关系,跳转位置,一般放在/WEB-INF目录下
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE struts-config PUBLIC "-//Apache Software Foundation//DTD Struts Configuration 1.3//EN" "http://struts.apache.org/dtds/struts-config_1_3.dtd">;
<struts-config>
  <!-- 配置表单 -->
  <form-beans>
      <!-- name是表单名字,可以随意写,但我们建议取名规范,表单为类名小写 -->
      <!-- type用于指定表单类全路径 -->
      <form-bean name="userForm" type="com.cakin.forms.UserForm"></form-bean>
  </form-beans>
  <action-mappings>
    <!-- 配置具体的一个action, path:表示将来访问action的url http://localhost:8080/web/path -->
      <!--  name用于关联某个表单-->
      <!--  type用于指定该action类全路径-->
      <action path="/login" name="userForm" type="com.cakin.actions.LoginAction">
      <!--  这里配置跳转关系-->
          <!--name表示结果名称 path:转发到哪个页面  -->
          <forward name="ok" path="/WEB-INF/wel.jsp"></forward>
          <forward name="err" path="/WEB-INF/err.jsp"></forward>
      </action>
      
  </action-mappings>
</struts-config>
 
六 写出welcome.jsp页面
<%@ page language="java" import="java.util.*" pageEncoding="ISO-8859-1"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
  <head>
    <base href="<%=basePath%>">
   
    <title>My JSP 'wel.jsp' starting page</title>
   
        <meta http-equiv="pragma" content="no-cache">
        <meta http-equiv="cache-control" content="no-cache">
        <meta http-equiv="expires" content="0">   
        <meta http-equiv="keywords" content="keyword1,keyword2,keyword3">
        <meta http-equiv="description" content="This is my page">
        <!--
        <link rel="stylesheet" type="text/css" href="styles.css">
        -->
 
  </head>
 
  <body>
    welcome <br>
  </body>
</html>
 
七 配置web.xml中的ActionServlet
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"; xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"; version="2.5" xsi:schemaLocation="http://java.sun.com/xml/ns/javaeehttp://java.sun.com/xml/ns/javaee/web-app_2_5.xsd";>
  <display-name />
  <servlet>
    <servlet-name>action</servlet-name>
    <servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
    <!-- 配置struts-config.xml -->
    <init-param>
      <param-name>config</param-name>
      <param-value>/WEB-INF/struts-config.xml</param-value>
    </init-param>
    <init-param>
      <param-name>debug</param-name>
      <param-value>3</param-value>
    </init-param>
    <init-param>
      <param-name>detail</param-name>
      <param-value>3</param-value>
    </init-param>
    <load-on-startup>0</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>action</servlet-name>
    <url-pattern>*.do</url-pattern>
  </servlet-mapping>
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
</web-app>
 
八 测试

 



 
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