0

我正在尝试检查用户当前是否已登录。如果用户已登录,我想将正确<li>的 ' 回显到菜单中。

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";
}
?>

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>
4

1 回答 1

0

您使用了错误的代码来检查用户是否登录,它是:isUserLoggedIn

尝试这个:

<?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";
}
?>
于 2016-02-27T03:43:30.277 回答