如何设置当前页面“活动”在php [英] How to set current page "active" in php
本文介绍了如何设置当前页面“活动”在php的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
您好,我在每个网页上有一个在我的网站上的菜单,我想把它在自己的menu.php文件,但我不知道如何设置 class =活动无论我在哪里。
这是我的代码:请帮助我
Hi I have a menu on my site on each page, I want to put it in it's own menu.php file but i'm not sure how to set the class="active" for whatever page i'm on. Here is my code: please help me
menu.php:
<li class=" has-sub">
<a class="" href="javascript:;"><i class=" icon-time"></i> Zeiten<span class="arrow"></span></a>
<ul class="sub">
<li><a class="" href="offnungszeiten.php">Öffnungszeiten</a></li>
<li><a class="" href="sauna.php">Sauna</a></li>
<li><a class="" href="frauensauna.php">Frauensauna</a></li>
<li class=""><a class="" href="custom.php">Beauty Lounge</a></li>
<li><a class="" href="feiertage.php">Feiertage</a></li>
</ul>
</li>
推荐答案
页面,并将其与当前活动页面一起传递到视图文件:
It would be easier if you would build an array of pages in your script and passed it to the view file along with the currently active page:
//index.php or controller
$pages = array();
$pages["offnungszeiten.php"] = "Öffnungszeiten";
$pages["sauna.php"] = "Sauna";
$pages["frauensauna.php"] = "Frauensauna";
$pages["custom.php"] = "Beauty Lounge";
$pages["feiertage.php"] = "Feiertage";
$activePage = "offnungszeiten.php";
//menu.php
<?php foreach($pages as $url=>$title):?>
<li>
<a <?php if($url === $activePage):?>class="active"<?php endif;?> href="<?php echo $url;?>">
<?php echo $title;?>
</a>
</li>
<?php endforeach;?>
使用 Smarty your menu.php会更美观:
With a templating engine like Smarty your menu.php would look even nicer:
//menu.php
{foreach $pages as $url=>$title}
<li>
<a {if $url === $activePage}class="active"{/if} href="{$url}">
{$title}
</a>
</li>
{/foreach}
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