1.求jsp登录源码 急急急急急急急急急急急
2.为什么不改动jsp中的管a管java代码,就不用重新构建
求jsp登录源码 急急急急急急急急急急急
登陆页面 index.jsp源码：
<%@ page language="java" import="java.util.*" pageEncoding="utf-8"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4. Transitional//EN">
<html>
<head>
<base href="<%=basePath%>">
<title>login</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="LoginServlet" method="post">
用户名：<input type="text" name="username" ><br>
密码:<input type="password" name="userpass"><br>
<input type="submit" value="登陆"> <input type="reset" value="取消">
</form>
</body>
</html>
-------------
LoginServlet.java 源码：
package servlet;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class LoginServlet extends HttpServlet {
/
*** Constructor of the object.
*/
public LoginServlet() {
super();
}
/
*** Destruction of the servlet. <br>
*/
public void destroy() {
super.destroy(); // Just puts "destroy" string in log
// Put your code here
}
/
*** The doGet method of the servlet. <br>
*
* This method is called when a form has its tag value method equals to get.
*
* @param request the request send by the client to the server
* @param response the response send by the server to the client
* @throws ServletException if an error occurred
* @throws IOException if an error occurred
*/
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
//获得jsp页面传输的参数
String username=request.getParameter("username");
String userpass=request.getParameter("userpass");
//判断
if(username.equals("user")&&userpass.equals("")){
response.sendRedirect("1.jsp");
}else if(username.equals("admin")&&userpass.equals("")){
response.sendRedirect("2.jsp");
}else{
response.sendRedirect("index.jsp");
}
}
/
*** The doPost method of the servlet. <br>
*
* This method is called when a form has its tag value method equals to post.
*
* @param request the request send by the client to the server
* @param response the response send by the server to the client
* @throws ServletException if an error occurred
* @throws IOException if an error occurred
*/
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
this.doGet(request, response);
}
/
*** Initialization of the servlet. <br>
*
* @throws ServletException if an error occurs
*/
public void init() throws ServletException {
// Put your code here
}
}
-------------
1.jsp:
<%@ page language="java" import="java.util.*" pageEncoding="utf-8"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4. Transitional//EN">
<html>
<head>
<base href="<%=basePath%>">
<title>My JSP '1.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>
This is 1.jsp <br>
</body>
</html>
-------------
2.jsp
<%@ page language="java" import="java.util.*" pageEncoding="utf-8"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4. Transitional//EN">
<html>
<head>
<base href="<%=basePath%>">
<title>My JSP '1.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>
This is 2.jsp <br>
</body>
</html>
为什么不改动jsp中的java代码,就不用重新构建
Tomcat会对你的的代码进行编译、编译，理源理系喂Capp源码 其实主要是码j码( 1110)B的源码对你 java代码进行解释编译，如果你不对java代码进行修改的统源代刷源码美化话是不用重新构建的，但是管a管排名统计源码php如果你对java代码进行了修改，那么就需要重新构建，理源理系索尼 kodi 源码输出如果不重新构建的码j码话，在执行的统源时候还是使用的原来未修改的代码！
管a管2024-12-22 11:52
2024-12-22 11:37
2024-12-22 11:35
2024-12-22 11:01
2024-12-22 10:54
2024-12-22 10:35
