if("ts".equals(name)&&"123456".equals(password)){
out.println("登录成功！");
request.getRequestDispatcher("index.jsp");
}else{
out.println("用户名或密码错误");
//request.getRequestDispatcher("Ajax/sale.html");
response.sendRedirect("sale.html");
//response.sendRedirect(request.getContextPath() + "/Registe.html");
//request.getRequestURI()/Ajax/Weibo.html
}
各种方法都试了
request.getRequestDispatcher(）根本不响应
求解，如何在验证用户名完成后直接跳转到一个新的jsp或者html文件
求大神解答。
out.println("登录成功！");
request.getRequestDispatcher("index.jsp");
}else{
out.println("用户名或密码错误");
//request.getRequestDispatcher("Ajax/sale.html");
response.sendRedirect("sale.html");
//response.sendRedirect(request.getContextPath() + "/Registe.html");
//request.getRequestURI()/Ajax/Weibo.html
}
各种方法都试了
request.getRequestDispatcher(）根本不响应
求解，如何在验证用户名完成后直接跳转到一个新的jsp或者html文件
求大神解答。
request.getRequestDispatcher(“/index.jsp”).forward(request,response);servlet页面跳转一般就是这两种
应该这样写
request.getRequestDispatcher("/a.jsp").forward(request, response);
<html>
<head>
<meta charset="UTF-8">
<title></title>
<link rel="stylesheet" type="text/css" href="css/normalize.css" />
<link rel="stylesheet" type="text/css" href="css/demo.css" />
<!--必要样式-->
<link rel="stylesheet" type="text/css" href="css/component.css" />
<script type="text/javascript" src="JS/jquery-1.11.0.js"></script>
<script type="text/javascript" src="JS/jquery-1.9.1.min.js"></script>
<script type="text/javascript" src="JS/prototype-1.6.0.3.js"></script>
<script type="text/javascript">
function getXhr() {
var xhr = null;
if(window.XMLHttpRequest) {
xhr = new XMLHttpRequest();
} else {
xhr = new ActiveXObject("Microsoft.XMLHttp");
}
return xhr;
}
function getText() { //1.获取Ajax对象
var xhr=getXhr();
//2.设置回调函数
xhr.onreadystatechange=function(){
if(xhr.readyState==4&&xhr.status==200){
var txt=xhr.responseText;
}
};
//3.创建请求
xhr.open("post","post_text.do",true);
xhr.setRequestHeader("content-type", "application/x-www-form-urlencoded");
//4.发送请求
xhr.send("uname=DR.");
} function check() { //1.获取Ajax对象
var xhr=getXhr();
//2.设置回调函数
xhr.onreadystatechange=function(){
if(xhr.readyState==4&&xhr.status==200){
document.getElementById("name_msg").innerHTML=xhr.responseText;
}
};
//3.创建请求
xhr.open("post","check_admin.do",true);
xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
document.getElementById("name_msg").innerHTML="正在检查....";
var uname=document.getElementById("logname").value;
var upassword=document.getElementById("logpass").value;
//4.发送请求
xhr.send("uname="+uname+"upassword="+upassword);
//window.location.href="http://localhost:8080/Ajax/Registe.html";
}
</script>
</head> <body>
<div class="container demo-1">
<div class="content">
<div id="large-header" class="large-header">
<canvas id="demo-canvas"></canvas>
<div class="logo_box">
<h3>欢迎来到我的微博</h3>
<form action="#" name="f" method="post">
<div class="input_outer">
<span class="u_user"></span>
<input name="logname" id="logname" class="text" style="color: #FFFFFF !important" type="text" placeholder="请输入账户">
</div>
<div class="input_outer">
<span class="us_uer"></span>
<input name="logpass" id="logpass" class="text" style="color: #FFFFFF !important; position:absolute; z-index:100;"value="" type="password" placeholder="请输入密码">
</div>
<div class="mb2"><a class="act-but submit" href="javascript:;" style="color: #FFFFFF" onclick="check();"/>登录</a></div>
<div class="mb2"><a class="act-but submit" href="javascript:;" style="color: #FFFFFF" onclick="liulan();">游客</a></div>
<span id="name_msg" style="color:red;"></span>
</form>
</div>
</div>
</div>
</div><!-- /container -->
<script src="JS/TweenLite.min.js"></script>
<script src="JS/EasePack.min.js"></script>
<script src="JS/rAF.js"></script>
<script src="JS/demo-1.js"></script>
</div>
</body></html>
求大佬解惑script里是根据老师代码照葫芦画瓢模仿的，求解惑。
response.setContentType("text/xml;charset=UTF-8");
PrintWriter out = response.getWriter();
out.write("登录成功！");
out.close();然后跳转页面要在前台回调方法里处理。要不就不用ajax，直接表单提交那就更简单
那如何调回前台，刚刚学习java还不太了解，上面代码的意思不就是输出登录成功四个字么？
document.getElementById("name_msg").innerHTML=xhr.responseText;
//用js操作浏览器跳转
}