我有这个代码似乎不工作,我想要做的是做一个查询来了解用户的级别,然后根据该级别我需要打印不同的内容,就目前它打印默认之一,我收到此错误信息:Php会话,根据用户级别问题重定向
mysql_result()预计参数1是资源,布尔在这条线下式给出:
$rank = mysql_result($rank1, 0, 'rank');
的代码是这样的:
session_start();
if(!$_SESSION['username']){
header("location:login.php"); // Redirect to login.php page
}
else //Continue to current page
header('Content-Type: text/html; charset=utf-8');
$rank1 = mysql_query("SELET access FROM tbl_galleries WHERE column='username" . mysql_real_escape_string($_SESSION['username']) . "'");
$rank = mysql_result($rank1, 0, 'rank');
switch ($rank)
{
case 3:
echo "<div><a href='#'>Order DIAMOND Membership</a></div>";
break;
case 2:
echo "<div><a href='#'>Order PLATINUM Membership</a></div>
<div><a href='#'>Order DIAMOND Membership</a></div>";
break;
case 1:
echo "<div><a href='#'>Order ELITE Membership</a></div>
<div><a href='#'>Order PLATINUM Membership</a></div>
<div><a href='#'>Order DIAMOND Membership</a></div>";
break;
default:
echo "<div><a href='#'>Order PRO Membership</a></div>
<div><a href='#'>Order ELITE Membership</a></div>
<div><a href='#'>Order PLATINUM Membership</a></div>
<div><a href='#'>Order DIAMOND Membership</a></div>";
}
也有错字“SELET”在 –