我使用XAMPP版本1.6.8创建了一个网站,然后安装了新版本的XAMPP 1.8.3 现在我的登录代码适用于1.6.8但不适用于1.8.3。
错误是
Parse error: syntax error, unexpected end of file in C:\xampp\htdocs\karimunjawa\admin\ceklogin.php on line 61
第61行是
</html>
ceklogin.php
<!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>Login</title>
<?php
//session_start();
include "../include/koneksi.inc";
if(isset($_POST['login'])) {
$username = $_POST['username'];
$password = $_POST['password'];
$sql = mysql_query("SELECT * FROM admin WHERE username='$username' && password='$password'");
$num = mysql_num_rows($sql);
if($num==1) {
// login benar //
//session_register("admin");
//$_SESSION['username'] = $username;
//$_SESSION['password'] = $password;
?><script language="JavaScript">alert('Anda berhasil login... Selamat datang <?php echo $_SESSION['username'] ?>..!!' );
document.location='adminfiles/'</script><?
} else {
// jika login salah
?><script language="JavaScript">alert('Username atau password Anda
salah'); document.location='index.php'</script><?
}
}
?>
<link rel="stylesheet" type="text/css" href="style.css" />
<style type="text/css">
<!--
.style2 {color: #000000}
-->
</style>
</head>
<body>
<div class="container">
<form action="" name="login" method="post">
<table width="279" height="110" border="0" align="center" cellspacing="0px">
<tr>
<td height="26" align="center" bgcolor="#993300"> <h3> Login Administrator </h3> </td>
</tr>
<tr>
<td height="32" align="center" bgcolor="#FFFFFF"><span class="style2">Username atau Password salah!!</span></td>
</tr>
<tr>
<td height="37" align="center" bgcolor="#FFFFFF"><a href="index.php"> Ulangi Login </a> </td>
</tr>
<tr>
<td align="center" bgcolor="#993300"> </td>
</tr>
</table>
</form>
</div>
</body>
</html>
请帮帮我 感谢