我正在尝试在用户访问该页面时为菜单添加class=active
。
我的菜单有以下代码:
<!-- main nav -->
<div class="header-nav navbar-collapse collapse">
<ul class="nav navbar-nav">
<li class="active"><a href="studenti.php">Acasa<i
class="fa fa-chevron-down"></i></a>
</li>
<li><a href="noutati.php">Anunturi si Noutati<i
class="fa fa-chevron-down"></i></a>
</li>
<li><a href="burse.php">Burse<i class="fa fa-chevron-down"></i></a>
</li>
<li><a href="orar.php">Orar<i class="fa fa-chevron-down"></i></a>
</li>
<li><a href="cazari.php">Cazari<i class="fa fa-chevron-down"></i></a>
</li>
<li><a href="javascript:;">Acces <i class="fa fa-chevron-down"></i></a>
<ul class="sub-menu">
<li><a href="#">Traiectorie scolara</a></li>
<li><a href="#">IDFR</a></li>
</ul>
</li>
</ul>
</div>
为了在访问页面时将链接菜单/页面突出显示为活动状态,我需要进行哪些设置?
我使用header.php作为全局标头配置。
答案 0 :(得分:0)
您可以通过拆分网址的末尾来检测网页:
var url = window.location;
var page = url.split('/');
page = page.slice(-1).pop();
然后只检测条件,例如
if(page == 'burse.php') {
$('.active').removeClass('active');
$('nav li:nth-child(3)').addClass('active');
}
答案 1 :(得分:0)
我试过这个并且它正在工作
<?php
function active($currect_page){
$url_array = explode('/', $_SERVER['REQUEST_URI']) ;
$url = end($url_array);
if($currect_page == $url){
echo 'active'; //class name in css
}
}
?>
<div class="header-nav navbar-collapse collapse">
<ul class="nav navbar-nav">
<li class="class="<?php active('studenti.php');?>><a href="studenti.php">Acasa<i class="fa fa-chevron-down"></i></a>
</li>
<li class="<?php active('noutati.php');?>"> <a href="noutati.php">Anunturi si Noutati<i class="fa fa-chevron-down"></i></a>
</li>
<li class="<?php active('burse.php');?>"> <a href="burse.php">Burse<i class="fa fa-chevron-down"></i></a>
</li>
<li class="<?php active('orar.php');?>"> <a href="orar.php">Orar<i class="fa fa-chevron-down"></i></a>
</li>
<li class="<?php active('cazari.php');?>"> <a href="cazari.php">Cazari<i class="fa fa-chevron-down"></i></a>
</li>
感谢你的回复!