TengFei’s Blog - java EE 简单登陆跳转逻辑

1、登陆页面

<!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>
    <!--把表单内容提交到myProject工程下的login-->
    <form action="/myProject/login" method="post">
        用户名： <input type="text" name="username" /><br>
        密  &nbsp;&nbsp;&nbsp;码：   <input type="password" name="password" /><br> 
        <input type="submit" value="登录" />
    </form>
</body>
</html>

2、login Servlet

        //获取用户名和密码
        String username =  request.getParameter("username");
        String password= request.getParameter("password");
        //在控制台打印
        System.out.println(username);
        System.out.println(password);
        
        //判断
        if(username.equals("admin") && password.equals("123")){
            response.sendRedirect("/myProject/welcom.html");
        }
        else{
            response.sendRedirect("/myProject/login.html");
        }