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简单例子分析struts的自定义拦截器

2016-08-09 19:37 281 查看

首先,创建web项目,配置好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" 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-auth</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.ng.filter.StrutsPrepareAndExecuteFilter</filter-class>
</filter>

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

</web-app>

然后,在webRoot目录下编写一个简单的前台页面:

<%@ page language="java" contentType="text/html; charset=utf-8"
pageEncoding="utf-8"%>
<%@taglib prefix="s" uri="/struts-tags" %>
<!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>
<h2>用户登录</h2>
<form action="admin.html" method="post">
用户名:
<input type="text" name="username">${loginError}
<input type="submit" value="登录">
</form>
</body>
</html>

在WEB-INF下新建webPage目录,建立一个简单的admin.jsp页面

<%@ 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>
后台管理页面!
</body>
</html>

建立Action类:

package com.auth.action;

import java.util.Map;

import org.apache.struts2.interceptor.SessionAware;

import com.opensymphony.xwork2.ActionSupport;

public class AdminAction extends ActionSupport implements SessionAware {
private String username;
private Map<String, Object> session;

public String getUsername() {
return username;
}

public void setUsername(String username) {
this.username = username;
}

public Map<String, Object> getSession() {
return session;
}

public void setSession(Map<String, Object> session) {
this.session = session;
}

public String login() {
if(username.equals("admin")) {
session.put("username", username);
return SUCCESS;
}else {
session.put("loginError", "用户名错误!");
return ERROR;
}
}
}

建立Inteceptor类:

package com.auth.interceptor;

import com.opensymphony.xwork2.ActionContext;
import com.opensymphony.xwork2.ActionInvocation;
import com.opensymphony.xwork2.interceptor.AbstractInterceptor;

public class UserInterceptor extends AbstractInterceptor {

@Override
public String intercept(ActionInvocation invocation) throws Exception {
String name = (String) ActionContext.getContext().
getSession().get("username");
if(name != null) {
String result = invocation.invoke();
return result;
}
return "error";
}

}

配置struts.xml文件:

<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE struts PUBLIC
"-//Apache Software Foundation//DTD Struts Configuration 2.1//EN"
"http://struts.apache.org/dtds/struts-2.1.dtd">

<struts>

<constant name="struts.devMode" value="true" />
<!-- action的后缀名为html -->
<constant name="struts.action.extension" value="html"></constant>

<package name="auth" extends="struts-default" namespace="/">
<interceptors>
<interceptor name="auth" class="com.auth.interceptor.UserInterceptor"></interceptor>
<interceptor-stack name="myStack">
<interceptor-ref name="auth"></interceptor-ref>
<interceptor-ref name="defaultStack"></interceptor-ref>
</interceptor-stack>
</interceptors>

<action name="admin" class="com.auth.action.AdminAction" method="login">
<result>/WEB-INF/webPage/admin.jsp</result>
<result name="error">/login.jsp</result>
</action>

<action name="auth">
<result>/WEB-INF/webPage/admin.jsp</result>
<interceptor-ref name="myStack"></interceptor-ref>
<result name="error">/login.jsp</result>
</action>
</package>
</struts>

然后,直接访问auth.html会跳转到登录页面,然后输入admin点击登录,正确跳转到后台页面,接着由于username已经存放在session中,所以再次访问auth.html能够正常跳转到admin.html的后台页面.

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