declaring session variables
Posted: Mon Aug 14, 2006 4:16 am
feyd | Please use
feyd | Please use
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,Code: Select all
and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]
im new to php. I have this login script that works fine just trying ot add a session variable to get the data from the field usertype (this will be set as admin or user)
ive tried the code below (used in main cod shown below) but getting parser errors the parts i have added that are not functioning are
$usertype = $_POST['usertype'];
$_SESSION['valid_type'] = get($usertype, "admin");
Just want to know how to get this session to work correctly like the others do any help appreciated
main code below.Code: Select all
<?
include "functions/include_fns.php";
if (isset($_POST['username']) && isset($_POST['password'])) {
$username = $_POST['username'];
$password = $_POST['password'];
$usertype = $_POST['usertype'];
if (login($username, $password, "admin")) {
$sql
$_SESSION['valid_user'] = $username;
$_SESSION['valid_name'] = getname($username, "admin");
$_SESSION['valid_type'] = get($usertype, "admin");
if (isset($_POST['url']) && !empty($_POST['url'])) {
$go_url = "".$url;
header("Location: $go_url");
} else {
header("Location: home.php");
}
exit;
} else {
//status = 0: Incorrect Username or password
//status = 1: Session has timed out or your are not logged in
header("Location: index.php?status=0");
exit;
}
} else {
header("Location: index.php?status=0");
exit;
}
?>feyd | Please use
Code: Select all
,Code: Select all
and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]