用php动态编写页面标题和活动类答案

【问题标题】：Dynamically writing page titles and active classes with php用php动态编写页面标题和活动类
【发布时间】：2010-05-07 19:55:55
【问题描述】：

一段时间以来，我一直在使用以下代码来动态写入 html 页面标题并为菜单项添加一个活动类。这仍然是实现这一目标的一个好方法，还是有更好/更智能/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' : '');?>

【问题讨论】：

标签： php html xhtml

【解决方案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" ';
    }
}

例如

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

【讨论】：

你甚至可以通过一个函数创建整行......

【解决方案2】：

这不是一个坏方法。我用循环来做，所以打字更少。例如：

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";
}

【讨论】：