ajax 登录问题 - Web 开发 / Ajax
login.html
<form action="" method="post"> 用户名: <input type="text" name="ad_user" id="ad_user" /><br /> 密码: <input type="password" name="ad_pass" id="ad_pass" /><br /> <input type="submit" name="submit" id="btn" value="提交" /></form>
$(function () { FormEve()})function FormEve(){ var $btn = $("#btn"); var $ad_user = $("#ad_user") var $ad_pass = $("#ad_pass") $btn.click(function(){ $.ajax({ type: "POST", url: "login.php", data: "ad_user="+ $ad_user.val() +"&ad_pass="+ $ad_pass.val(), success: function(msg){ alert( "Data Saved: " + msg ); } }); })}<?php session_start(); //注销登录if($_GET['action'] == "logout"){unset($_SESSION['userid']);unset($_SESSION['username']);header("Location:login.html");exit;}//登录if(!isset($_POST['submit'])){exit('非法访问!');}$ad_user = htmlspecialchars($_POST['ad_user']);$ad_pass = MD5($_POST['ad_pass']);//数据库连接include('conn.php');//检测用户名及密码是否正确$check_query = mysql_query("select adminid from admin where ad_user='$ad_user' and ad_pass='$ad_pass' limit 1");if($result = mysql_fetch_array($check_query)){//登录成功$_SESSION['username'] = $ad_user;$_SESSION['userid'] = $result['adminid'];header("Location:index.php");die('Could not connect: ' . mysql_error());exit;} else {header("Location:login.html");}?><?php session_start(); ?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html"; charset="utf-8" /><title></title><link href="css/admin_dig.css" rel="stylesheet" type="text/css" /><link href="css/common.css" rel="stylesheet" type="text/css" /></head><body><?phpif(!isset($_SESSION['userid'])){ header("Location:login.html"); exit();}?><?php echo "".$_SESSION['username']."" ?><a href="login.php?action=logout">注销</a></body></html>