这段代码什么地方有问题？

蛋定

这段代码哪里有问题哦？怎么我一直实现不了，求高手解决！

1. <?php
   
2. header ( "content-type:text/html; charset=utf-8" );
   
3. 
   
4. @mysql_connect ( '218.104.***.145', 'sq8******', '*******' )or die("数据库连接出错！");
   
5. @mysql_select_db ( 'sq8******' ) or die("数据表选择出错！");
   
6. mysql_query ( "set names 'utf8'" );
   
7. $result = mysql_query ( "select ip from ip_sheet" );
   
8. 
   
9. $visit_ip = GetIP (); //获取访问者ip地址
   
10. while ($ip = mysql_fetch_array($result)){
    
11. 
    
12.    if($ip['ip'] == $visit_ip){
    
13.    
    
14.      echo("<script language='javascript'>location='skipPage/error.html';</script>");
    
15.    
    
16.    }else{
    
17.    
    
18.     fopen("indexa.php");
    
19.    
    
20.    }
    
21. 
    
22. }
    
23. /*
    
24. * 获取访问者ip地址函数
    
25. *
    
26. */
    
27. function GetIP() {
    
28. if (! empty ( $_SERVER ["HTTP_CLIENT_IP"] )) {
    
29. $cip = $_SERVER ["HTTP_CLIENT_IP"];
    
30. } elseif (! empty ( $_SERVER ["HTTP_X_FORWARDED_FOR"] )) {
    
31. $cip = $_SERVER ["HTTP_X_FORWARDED_FOR"];
    
32. } elseif (! empty ( $_SERVER ["REMOTE_ADDR"] )) {
    
33. $cip = $_SERVER ["REMOTE_ADDR"];
    
34. } else {
    
35. $cip = "无法获取！";
    
36. }
    
37. return $cip;
    
38. }
    
39. 
    
40. ?>

复制代码