我试图检查用户当前是否已登录。
如果用户已登录,我想在菜单中回显正确的<li>
。
我尝试在页面顶部执行此操作:
<?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";
}
?>
以及<li>
代码:
<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 $r2 ?></a></li>
答案 0 :(得分:0)
您使用错误的代码进行检查用户是否已登录,它是: isUserLoggedIn
试试这个:
<?php
// Report all PHP errors (see changelog)
error_reporting(E_ALL);
require_once ('models/config.php');
//$username = $loggedInUser->username;
if (isUserLoggedIn()) {
$r1 = $loggedInUser->username;
$r2 = "Logout";
} else {
$r1 = "Login";
$r2 = "Register";
}
?>