基于Struts 二 Ajax实现的Login应用

2012-09-19

基于Struts 2 Ajax实现的Login应用Struts 2内嵌了Dojo工具包，实现对Ajax的支持。下面是一个用户名和密码都

基于Struts 2 Ajax实现的Login应用

Struts 2内嵌了Dojo工具包，实现对Ajax的支持。下面是一个用户名和密码都是Admin的Login应用。

?1、在struts.xml中加入一个Action mapping

<action?name="showAjaxLoginForm">?? ???????<result>/pages/ajaxlogin.jspresult>??
action>?? ??
<action?name="ajaxLogin"?class="net.roseindia.Login">?? ???????<result?name="input">/pages/ajaxlogin.jspresult>??
???????<result?name="error">/pages/ajaxlogin.jspresult>?? ???????<result>/pages/ajaxloginsuccess.jspresult>??
>?

?2、用Ajax编写一个Login页面ajaxlogin.jsp

?这个页面使用了标签, 这个标签能通过Ajax tags载入页面内容。jsp页面还使用了标签，这个标签可以利用Ajax来更形页面元素和提交一个form。当出现错误是，和标签执行并显示错误信息。

xml 代码?

<%@?taglib?prefix="s"?uri="/struts-tags"%>?? <html>??
??<head>?? ????<s:head?theme="ajax"?debug="true"/>?? ??head>??
??<body>?? ????<s:div?id="loginDiv"?theme="ajax">??
????<div?style="width:?300px;border-style:?solid">?? ??????<s:form?action="ajaxLogin"??validate="true">??
????????<tr>?? ??????????<td?colspan="2">??
????????????Login ?? ??????????td>??
????????tr>?? ????????<tr>??
??????????<td?colspan="2">?? ????????????<s:actionerror?/>??
????????????<s:fielderror?/>?? ??????????td>??
????????tr>?? ?? ??????????<s:textfield?name="username"?label="Login?name"/>??
??????????<s:password?name="password"?label="Password"/>?? ??????????<s:submit?theme="ajax"?targets="loginDiv"?notifyTopics="/ajaxLogin"/>???? ??
??????s:form>?? ????div>??
????s:div>?? ??body>??
html>???

?3、编写一个验证用户名和密码的Action类Login.java

如果验证成功返回SUCCESS，失败就返回ERROR

java 代码?

package?net.roseindia; ?? ?? import?com.opensymphony.xwork2.ActionSupport; ??
import?java.util.Date; ?? ?? /** ?
?*?Validate?a?user?login.?? ?*/??
public?class?Login?extends?ActionSupport?{ ?? ??
????public?String?execute()?throws?Exception?{ ?? ????????System.out.println("Validating?login?...?..."); ??
????????System.out.println("User?=?"?+?getUsername()); ?? ????????if?(!getUsername().equals("Admin")?||?!getPassword().equals("Admin"))?{ ??
????????????System.out.println("Validating?error?!?User?=?"?+?getUsername()); ?? ????????????addActionError("Invalid?user?name?or?password!?Please?try?again!"); ??
????????????return?ERROR; ?? ????????}?else?{ ??
????????????System.out.println("Validating?success?!"); ?? ????????????return?SUCCESS; ??
????????} ?? ????} ??
?? ????//?----?Username?property?---- ??
?? ????/** ?
?????*?Field?to?store?User?username.? ?????*/??
????private?String?username?=?null; ?? ??
????public?String?getUsername()?{ ?? ????????return?username; ??
????} ?? ??
????public?void?setUsername(String?value)?{ ?? ????????username?=?value; ??
????} ?? ??
????//?----?Username?property?---- ?? ??
????/** ? ?????*?Field?to?store?User?password.?
?????*/?? ????private?String?password?=?null; ??
?? ????public?String?getPassword()?{ ??
????????return?password; ?? ????} ??
?? ????public?void?setPassword(String?value)?{ ??
????????password?=?value; ?? ????} ??
?? }???

?4、编写一个登录成功页面ajaxloginsuccess.jsp

xml 代码

<html>?? ??<head>??
????<title>Login?Successtitle>?? ??head>??
??<body>?? ????<p?align="center"><font?color="#000080"?size="5">Login?Successful?!font>p>??
????<h1>?Welcome?to?<%=request.getParameter("username")%>??h1>?? ??body>??
html>???

5、访问下面连接 http://localhost:8080/s2ajax/showAjaxLoginForm.action

来自：http://www.iteye.com/topic/124516

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