Php determine whether a user is logged on _ PHP Tutorial

Php checks whether a user is logged on. To determine whether a user logs in, the first step is to save the data to the session after the user logs in successfully, then, when the user accesses the user permission that needs to be logged in, it determines whether the session determines whether the user logs in. the first step is to save the data to the session after the user logs in successfully, then, when the user accesses the user permission that needs to be logged on, it determines whether the session is empty. if not, the user logs on successfully. Next let's look at an instance.

To determine whether a user logs in, the first step is to save the data to the session after the user logs in successfully, then, when the user accesses the user permission that needs to be logged on, it determines whether the session is empty. if not, the user logs on successfully. Next let's look at an instance.

Session_start ();

If (getconfig ("chatroom_admin") == _ post ["username"] & getconfig ("chatroom_adminpassword") = md5 ($ _ post ["password"])
{
// Log in successfully and register the session
Session_register ("administrator ");
If (isset ($ _ session ["hack_num"])
{
Session_unregister ("hack_num ");
}
$ _ Session ["administrator"] = "yes ";
Header ("location: main. php ");
Exit;
}
Else
{
If ($ _ session ["hack_num"] = "")
{
Session_register ("hack_num ");
$ _ Session ["hack_num"] = 1;
}
Else
{
$ _ Session ["hack_num"] ++;

}
Header ("location:.../notice. php? Id = ". admin_login_lost );
Exit;
}
?>

If there is no red icon, the user logs on successfully and saves the data to the session ['hack _ num.

// This Page first checks whether the user is logged on correctly. if the user is not logged on, the logon page is displayed.
Session_start ();
If ((! $ User_id) or (! $ User_array [hack_num]) or (! $ User_array [hack_num]) {
Header ("location: login. php3 ");
}
?>

The above code is used to determine whether the session ['hack _ num'] has a value. If yes, the logon is successful.

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