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Тема |
Re: Личи си, че не е копнато [re: ro6avia] |
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Автор | Freak (Нерегистриран) | |
Публикувано | 24.01.07 14:50 |
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И скрипта ми изглежда така:
include_once('bookmark_fns.php');
session_start();
$username = $HTTP_POST_VARS['username'];
$password = $HTTP_POST_VARS['passwd'];
$id = $HTTP_SESSION_VARS['userid'];
if ($username && $password)
{
if (login($username, $password))
{
$HTTP_SESSION_VARS['valid_user'] = $username;
}
else
{
include_once('header.php');
echo '<br />';
echo '<center>';
echo 'Íå ìîæå äà áúäåòå ëîãíàò.
Òðÿáâà äà ñòå ëîãíàò, çà äà âèäèòå òàçè ñòðàíèöà.';
do_html_url('login.php', 'Login');
echo '</center>';
include_once('footer.php');
exit;
}
}
include_once('header.php');
echo '<center>';
check_valid_user();
get_user_pictures($HTTP_SESSION_VARS['valid_user']);
$conn = db_connect();
if (!$conn)
return 'Íÿìà âðúçêà ñ áàçàòà äàííè, ìîëÿ îïèòàéòå ñëåä ìàëêî.';
$query = "select rights from user where username=".$HTTP_SESSION_VARS['valid_user']."
and passwd='$password'";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
switch($row['rights'])
{
case '1':
display_user_menu();
break;
case '2':
display_user_menu_upload();
break;
case '3':
display_user_menu_delete();
break;
}
echo '</center>';
include_once('footer.php');
?>
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