我有包含在所有页面中的带有菜单的页眉。问题是当我单击子菜单时,我无法创建子菜单,并且它的父菜单突出显示或处于事件状态。
$(".dropdown .dropdown-menu li a").click(function() {
$(".active").removeClass('active');
$(this).closest('.dropdown').find('a:first').addClass("active");
$(this).addClass("active");
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<ul>
<li class="dropdown">
<a class="dropdown-toggle" href="#" id="builders">Builders<i class="fa fa-caret-down" aria-hidden="true"></i></a>
<ul class="dropdown-menu">
<li><a href="gurgaon.php">Gurgaon</a></li>
<li><a href="<?php echo base_url();?>all-india" class="active">All India</a></li>
<li><a href="gurgaon-projects.php">Gurgaon Projects</a></li>
<li><a href="<?php echo base_url();?>home/property_list">Property List</a></li>
</ul>
</li>
</ul>
最佳答案
您的 jQuery 代码可以运行,您只需要让它在加载时运行以及 click
,您可以使用 trigger('click')
来实现:
var $a = $(".dropdown .dropdown-menu li a").click(function(e) {
e.preventDefault(); // just for testing...
$(".active").removeClass('active');
$(this).closest('.dropdown').find('a:first').addClass("active");
$(this).addClass("active");
});
$a.filter('.active').trigger('click');
.active {
color: #C00;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<ul>
<li class="dropdown">
<a class="dropdown-toggle" href="#" id="builders">Builders<i class="fa fa-caret-down" aria-hidden="true"></i></a>
<ul class="dropdown-menu">
<li><a href="gurgaon.php">Gurgaon</a></li>
<li><a href="<?php echo base_url();?>all-india" class="active">All India</a></li>
<li><a href="gurgaon-projects.php">Gurgaon Projects</a></li>
<li><a href="<?php echo base_url();?>home/property_list">Property List</a></li>
</ul>
</li>
<li class="dropdown">
<a class="dropdown-toggle" href="#" id="builders">Builders<i class="fa fa-caret-down" aria-hidden="true"></i></a>
<ul class="dropdown-menu">
<li><a href="gurgaon.php">Gurgaon</a></li>
<li><a href="<?php echo base_url();?>all-india">All India</a></li>
<li><a href="gurgaon-projects.php">Gurgaon Projects</a></li>
<li><a href="<?php echo base_url();?>home/property_list">Property List</a></li>
</ul>
</li>
</ul>
关于javascript - 单击子项或单击子菜单时需要激活父类,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53222434/