正在尝试显示用户'；用户登录后，导航栏中的s用户名 - Trying to display the User's username in my Navigation Bar once user is logged in?

Trying to display the User's username in my Navigation Bar once user is logged in?

本文关键字：用户导航显示登录 | 更新日期: 2023-09-27

我在这段PHP中遇到语法错误：

<?php
if(!empty($_SESSION['user']))
{
    echo '<li class="dropdown">';
    Problem somewhere in here -> echo '<a href="#" class="dropdown-toggle" data-toggle="dropdown">Logged in ('echo htmlentities($_SESSION['user']['username'], ENT_QUOTES, 'UTF-8')')<span class="caret"></span></a>'; <-Problem somewhere in here
    echo '<ul class="dropdown-menu" role="menu">';
    echo '<li class="inactive"><a href="account.php">Account</a></li>';
    echo '<li class="inactive"><a href="logout.php">log out</a></li>';
    echo '</ul>';
}
    else{echo '<li class="inactive"><a href="login.php">Log In</a></li>';
}
?>

希望有比我更好的人能发现问题的原因？

它在Bootstrap下拉菜单中。如果你需要更多信息，我会尽力回答。谢谢

这是因为您在回显中回显。你应该像这样把字符串串起来：

echo '<a href="#" class="dropdown-toggle" data-toggle="dropdown">Logged in (' . htmlentities($_SESSION['user']['username'], ENT_QUOTES, 'UTF-8') . ')<span class="caret"></span></a>';

我建议使用更复杂的方法：

function returnContentUserSignedIn($user) {
        return '
        <li class="dropdown">
                <!-- CONTENT -->
                <a href="#" class="dropdown-toggle" data-toggle="dropdown">
                        Logged in ' . htmlspecialchars($user) . '<span class="caret"></span>
                </a>
                <!-- CONTENT -->
        </li>';
}
function returnContentUserSignedOut() {
        return '
        <li class="inactive">
                <!-- CONTENT -->
        </li>';
}
<?php echo ( !empty($_SESSION['user']) ? returnContentUserSignedIn($_SESSION['user']) : returnContentUserSignedOut() ); ?>