4

我有这段代码,我已经尝试了我能想到的一切,让它在我的 WAMP 本地服务器上工作,任何帮助将不胜感激。我是 PHP 笨蛋。这适用于实时服务器,但不适用于我的 WAMP 服务器。我确实登录了,只是页面似乎没有将会话变量传递给正确的用户级别。这就是第一次对糟糕的描述感到抱歉的原因。

if(!empty($_SESSION['LoggedIn']) && !empty($_SESSION['login']))
 {
if ($level == "Administrator") {

echo 'My Content';
}

elseif ($level == "Bank Officer") {
echo "";
}

elseif ($level == "Agent") {
echo "";
 }

elseif(!empty($_POST['login']) && !empty($_POST['password']))
{
$login = mysql_real_escape_string($_POST['login']);
$password = $_POST['password'];


$checklevel = mysql_query("SELECT * FROM users WHERE login = '".$login."' AND password = '".$password."' ");
if(mysql_num_rows($checklevel) == 1)
{
    $row = mysql_fetch_array($checklevel);
    $level = $row['level'];
    $_SESSION['level'] = $level;
}

$checklogin = mysql_query("SELECT * FROM users WHERE login = '".$login."' AND password = '".$password."' AND level='".$level."'");
if(mysql_num_rows($checklogin) == 1)
{
    $row = mysql_fetch_array($checklogin);
    $firstname = $row['firstname'];
    $login = $row['login'];
    $agent = $row['agent'];
    $_SESSION['agent'] = $agent;
    $_SESSION['firstname'] = $firstname;
    $_SESSION['login'] = $login;
    $_SESSION['LoggedIn'] = 1;

感谢您的任何帮助。

4

1 回答 1

2
if ($_SESSION['level'] == "Bank Officer") 
{
    header('Location: index3.php'); 
    exit;
}
elseif ($_SESSION['level'] == "Agent")
{
    header('Location: index4.php'); 
    exit;
}
elseif ($_SESSION['level'] == "Bank Manager") 
{
    header('Location: index5.php'); 
    exit;
}
else
{
    echo "Contact Administrator";
    exit;
}
于 2012-12-01T06:25:36.397 回答