ajax实现登录练习 

 1 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
 2 <html xmlns="http://www.w3.org/1999/xhtml">
 3 <head>
 4 <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
 5 <title>无标题文档</title>
 6 <script src="jq.js"></script> 
 7 </head>
 8 
 9 <body>
10 <h1>登录</h1>
11 
12 <div>用户名：</div>
13 <input type="text" id="uid" />
14 <div>密码：</div>
15 <input type="text" id="pwd" />
16 <input type="button" value="确定" id="dian" />
17 
18 </body>
19 </html>
20 <script type="text/javascript">
21 
22 $(document).ready(function(e) {
23     $("#dian").click(function(){
24         
25         var uid=$("#uid").val();
26         var pwd=$("#pwd").val();
27         
28         $.ajax({
29             
30             url:"chuli.php",
31             data:{u:uid,p:pwd},
32             type:"POST",
33             dataType:"TEXT",
34             success: function(data){
35             
36             if(data=="ok")
37             {
38                 window.location="Main.php";
39                 
40             }
41             else
42             {
43              alert(data);    
44             }
45             
46             } 
47         
48         });
49         
50         
51     })
52     
53 });
54 
55 
56 
57 
58 </script>
 1 <?php
 2 
 3 $uid=$_POST["u"];
 4 $pwd=$_POST["p"];
 5 
 6 include("DBDA.php");
 7 
 8 $db=new DBDA();
 9 
10 $sql=" select count(*) from login where username='{$uid}'and password='{$pwd}'";
11 $attr=$db->Query($sql);
12 if($attr[0][0]==0)
13 {
14     
15     echo "用户名密码不正确";    
16 }
17 else
18 {
19  echo "ok";    
20 }

 

posted on 2016-05-02 13:56  宁静致远-zw  阅读(495)  评论(0)    收藏  举报