Php I';我在从数据库显示此值时遇到问题
…同时连接两个表变量以供使用 以下是我要查询的内容:Php I';我在从数据库显示此值时遇到问题,php,arrays,login,Php,Arrays,Login,…同时连接两个表变量以供使用 以下是我要查询的内容: //GRABBING USER ID if (isset($_SESSION['login_email'])) { $session_entry = $_SESSION['login_email']; $user_query = mysqli_query($connection, "SELECT id FROM users WHERE email = '$session_entry'"); $user_assoc
//GRABBING USER ID
if (isset($_SESSION['login_email'])) {
$session_entry = $_SESSION['login_email'];
$user_query = mysqli_query($connection, "SELECT id FROM users WHERE email = '$session_entry'");
$user_assoc = mysqli_fetch_assoc($user_query);
$user_id = $user_assoc;
}
//GRABING ACCESS ID
if (isset($_SESSION['login_email'])) {
$access_query = mysqli_query($connection, "SELECT access FROM user_access WHERE user_id = '$user_id'");
$access_assoc = mysqli_fetch_assoc($access_query);
$access = $access_assoc;
}
还有$access echo的“数组”。。。我不知道该怎么办
现在,在我的user_access表中有两列:1。用户id和2。通路
我想让它回显访问代码,但正如上面所说,我得到的只是“数组”
我参与使用此代码的目的是为用户提供访问特定网页的访问代码,他们有权通过这些代码访问这些网页
$access = $access_assoc;
一定是
$access = $access_assoc['access'];
$user_id = $user_assoc['id'];
也
一定是
$access = $access_assoc['access'];
$user_id = $user_assoc['id'];
Checkmysqli\u fetch\u assoc始终返回
fieldname=>值的关联数组