Php 用户权限-子菜单
我正在尝试进行查询,以获取用户可以访问的子菜单 但我有一些调查 我创建此查询是为了获取菜单和子菜单Php 用户权限-子菜单,php,Php,我正在尝试进行查询,以获取用户可以访问的子菜单 但我有一些调查 我创建此查询是为了获取菜单和子菜单 $query = ("SELECT menu_id, menu_friendlyname, name FROM menu WHERE back = 1 AND active = 1"); $menu = db_array($query, '++-'); //function that i created to facilitate my life, ign
$query = ("SELECT menu_id, menu_friendlyname, name
FROM menu
WHERE back = 1 AND active = 1");
$menu = db_array($query, '++-'); //function that i created to facilitate my life, ignore
现在我试图只获取用户有权访问的子菜单
$query = ("SELECT * FROM menu m LEFT JOIN submenu s USING(menu_id) WHERE s.active = 1 AND m.active = 1 AND m.back = 1 AND s.submenu_id IN ('". $submenu_ids ."')");
$submenu = db_array($query, '+++-');
我正在从会话中获取子菜单ID
$submenu_ids = $_SESSION['user']['permissions_filter']['submenu_ids'];
我对这个问题做了回应,似乎效果不错
SELECT * FROM menu m LEFT JOIN submenu s USING(menu_id) WHERE s.active = 1 AND m.active = 1 AND m.back = 1 AND s.submenu_id IN ('1,3,5,6')
我做错了什么
谢谢。有错误消息吗?问题是什么?我已经解决了我的错误,我的问题是读“1,3,5,6”,而不是1,3,5,6,但是谢谢Matthew