我在两个表（login_pre，特权）之间创建关系，它们如下：
权限表：
+----------------------------------+
| AccessLevel | login_id | pre_id|
|----------------------------------|
| 1 | 1 | 1 |
| 2 | 1 | 2 |
| 4 | 2 | 4 |
+----------------------------------+
这是login_pre表：
+----------------------------------+
| username| userpass | login_id |
|----------------------------------|
| a | 123 | 1 |
| a | 123 | 1 |
| b | 1234 | 2 |
+----------------------------------+
其中添加访问权限级别的管理员
如何使读取访问级别=（1或2）的PHP代码login_id=1
我写了这个两页的代码，但这个只读访问级别 =2 的 login_id =1 ：
权限页面：
ob_start();
session_start();
include 'C:'xampp'htdocs'database'agtdatabase'agt_site'connection'connect.php';
$query ="SELECT * FROM privilege " ;
$result = mysqli_query($link,$query) or die('');
if(isset($_SESSION['sessionloginid']))// point to id of user logged in
{
$query ="SELECT * FROM privilege AS privilege.login_id JOIN login_pre AS login_pre.login_id ON privilege.login_id=login_pre.login_id WHERE login_id =1 " ;
$result = mysqli_query($link,$query) or die('');
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
// $_SESSION['sessionloginid']= $row['login_id'];// output 11
$_SESSION['sessionaccess']= $row['AccessLevel'];//output 12 for user login id = 1
}
}
ob_end_flush();
和保护页面：
<?php
include_once('C:'xampp'htdocs'database'agtdatabase'agt_site'login2'privilege.php');
function protect_page()
{
if($_SESSION['sessionloginid']== true && $_SESSION['sessionaccess'] )
{
//header ('location:http://localhost/database/agtdatabase/agt_site/agtSite/agt2.php');
$_SESSION['sessionloginid']; // output 1
echo $_SESSION['sessionaccess'] ;//output 2 for user loginid = 1
exit();
}
}
protect_page();
?>
单个查询怎么样：
SELECT * FROM privilege AS p JOIN login_pre AS l
ON p.login_id = l.login_id
WHERE AccessLevel IN (1, 2) AND l.login_id = 1;
更新
不确定您是否知道，但这部分始终覆盖会话数据，因此您只能获得最后一行：
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$_SESSION['sessionloginid']= $row['login_id'];
$_SESSION['sessionaccess']= $row['AccessLevel'];
}