我运行了这个脚本,但我一直收到我放入其中的“无效”消息。我知道数据库中的值是正确的,它一定是我的 PHP 中的东西......
$getadmin = "SELECT role FROM user WHERE user_id=$uid";
$showadmin = @mysqli_query ($dbc, $getadmin); // Run the query.
$admin = mysqli_fetch_assoc($showadmin);
if($admin == 'admin'){
echo 'You are an admin!';
} else {
echo 'Did not work';
}
最佳答案
试试这个:
$getadmin = "SELECT role FROM user WHERE user_id=$uid";
$showadmin = @mysqli_query ($dbc, $getadmin); // Run the query.
$role = mysqli_fetch_assoc($showadmin);
if($role['role'] == 'admin'){
echo 'You are an admin!';
} else {
echo 'Did not work';
}
关于PHP:如何运行此 IF 语句,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5822693/