我正在尝试检查用户当前是否登录。如果用户已登录,我希望将正确的<li>
返回到菜单。
我试着在页面顶部这样做:
<?php
// Report all PHP errors (see changelog)
error_reporting(E_ALL);
require_once ('models/config.php');
//$username = $loggedInUser->username;
if ($isUserLoggedIn()) {
$r1 = $loggedInUser->username;
$r2 = "Logout";
} else {
$r1 = "Login";
$r2 = "Register";
}
?>
这是<li>
标签:
<li><a class="short" href="About" style="display: block;"><?php echo $r1 ?></a></li>
<li><a class="short" href="About" style="display: block;"><?php echo $r2 ?></a></li>
您使用了错误的代码来检查用户是否登录,它是:isUserLoggedIn
试试这个:
<?php
// Report all PHP errors (see changelog)
error_reporting(E_ALL);
require_once ('models/config.php');
//$username = $loggedInUser->username;
if (isUserLoggedIn()) {
$r1 = $loggedInUser->username;
$r2 = "Logout";
} else {
$r1 = "Login";
$r2 = "Register";
}
?>