Question

一段时间以来，我一直在使用以下代码动态编写html页面标题，并为菜单项添加活动类。这仍然是一个很好的原因，为什么要实现这一点，还是有更好/更聪明/最佳的方法来实现同样的目标？

<?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='contact.php'? 'class="active"' : '');?>

菜单示例

<ul id="nav">
<li><a href="index.php"  target="_parent" <?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='index.php'? 'class="active"' : '');?>><span>Home</span></a></li>
<li><a href="services.php"  target="_parent" <?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='services.php'? 'class="active"' : '');?><?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='services-landlords.php'? 'class="active"' : '');?><?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='services-sellers.php'? 'class="active"' : '');?><?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='services-tennants.php'? 'class="active"' : '');?>><span>Our Services</span></a></li>
<li><a href="for-sale.php" target="_parent" <?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='sales.php'? 'class="active"' : '');?>><span>Sales</span></a></li>
<li><a href="to-let.php" target="_parent" <?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='lettings.php'? 'class="active"' : '');?>><span>Lettings</span></a></li>
<li><a href="register.php"  target="_parent" <?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='register.php'? 'class="active"' : '');?><?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='register-thanks.php'? 'class="active"' : '');?>><span>Register</span></a></li>
<li><a href="contact.php"  target="_parent" <?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='contact.php'? 'class="active"' : '');?>><span>Contact Us</span></a></li>
</ul>

页面标题示例

<?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='services.php'? 'Services' : '');?>
<?php echo (basename($_SERVER['SCRIPT_FILENAME'])=='services-landlords.php'? 'Services for Landlords' : '');?>

Answer 1

你可以把它放到一个函数中。

function menuIsActive ($filename)
{
    echo (basename($_SERVER['SCRIPT_FILENAME']) == $filename)
    {
        echo ' class="active" ';
    }
}

离。

<li><a href="contact.php" target="_parent" <?php menuIsActive("contact.php"); ?>>Contact Us</a></li>

Answer 2

这不是一个糟糕的方式。我用一个循环这样做，所以打字更少。 e.g：

foreach(array($pagenames as $pagename=>$pageaddress) {
  $active= $_SERVER('SCRIPT_FILENAME'])==$pageaddress? 'class="active"' : '';
  echo <li><a href="$pageaddress"  $active target="_parent">$pagename</a></li>\n";
}

使用php动态编写页面标题和活动类

2 个答案: