动态使侧边栏菜单处于活动状态

Question

我有一个包含侧边栏菜单项的母版页，footer.I 正在其他侧边栏中使用此侧边栏 pages.After 我登录仪表板菜单突出显示它是因为我正在制作它 active.But单击其他链接后，我怎样才能使该菜单处于活动状态，因为我使用的是所有 pages.Is 通用的侧边栏，有什么办法可以动态地做到这一点？或者我是否必须在每个页面中复制以下代码并激活特定的菜单项？

    <div class="col-xs-12 col-sm-12 col-md-12">
    <div class="LeftMenu">
    <ul>
    <a href="#!/admindashboardhome" title="Dashboard"><li 
    class="Active"><span class="glyphicon glyphicon-th"></span>
    <span>Dashboard</span></li></a>
    <a href="#!/examinationhalltickets" title="Declaration"><li><span
    class="glyphicon glyphicon-barcode"></span><span>Examination
    Form</span></li></a>
    <a href="#!/collegedetails" title="Declaration"><li><span 
    class="glyphicon glyphicon-education"></span><span>College 
    Details</span></li></a>
    <a href="#!/studentdetails" title="Declaration">
    <li><span class="glyphicon glyphicon-user"></span><span>Student
    Details</span></li></a>
    </ul>
    </div>
    </div>

Answer 1

这应该可以解决问题

HTML:

<a ng-class="getClass('/admindashboardhome')" href="/admindashboardhome">Dashboard</a>

JavaScript:

$scope.getClass = function (path) {
  return ($location.path().substr(0, path.length) === path) ? 'active' : '';
}

CSS:

.active {
  //styling
}

动态使侧边栏菜单处于活动状态

Dynamically making side bar menu as active

sidebar

angularjs