如何从php页面跳转到html页面_随笔

如何从php页面跳转到html页面

PHP页面首先要是在服务器里运行，即不能像Html页面一样直接双击查看，而是要在浏览器里输入相关地址才能访问；其次，PHP代码被执行的文件，必须是以.php结尾，不能是在.html结尾；再次，PHP代码有起始标签 <?php ?>标签不能少，少了它就当Html代码认了

html静态页面是不能接收参数的，你可以把html页面改为jsp页面，实现如下servlet类package one.twoimport java.io.IOException

import java.io.PrintWriterimport javax.servlet.RequestDispatcher

import javax.servlet.ServletException

import javax.servlet.http.HttpServlet

import javax.servlet.http.HttpServletRequest

import javax.servlet.http.HttpServletResponsepublic class Servlet extends HttpServlet {

public Servlet() {

super()

}/** 好网： http://www.ablanxue.com

* Destruction of the servlet. <br>

public void destroy() {

super.destroy()// Just puts "destroy" string in log

// Put your code here

} public void doGet(HttpServletRequest request, HttpServletResponse response)

throws ServletException, IOException {

doPost( request, response)

} public void doPost(HttpServletRequest request, HttpServletResponse response)

throws ServletException, IOException {request.setAttribute("param"," http://www.ablanxue.com")

RequestDispatcher requestDispatcher=request.getRequestDispatcher("/ablanxue.jsp")

requestDispatcher.forward(request,response)

} /**

* Initialization of the servlet. <br>

* @throws ServletException if an error occurs

public void init() throws ServletException {

// Put your code here

}}

jsp页面接收参数 ablanxue.jsp代码如下<%@ 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 4.01 Transitional//EN">

<html>

<head>

<title>My JSP 'ablanxue.jsp' starting page</title>

<!--

--> </head>

<body>

String PAR=(String)request.getAttribute("param")

上面这个PAR就是得到的参数了输入出

<%=PAR%>

页面会输出： http://www.ablanxue.com

<br>

</body>

</html>

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如何从php页面跳转到html页面

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