我在yii2中使用菜单小部件来创建侧边菜单。 我的菜单包含两个指向相同操作的菜单项。一个链接用于参数,另一个链接用于没有参数。 但是,当我点击任何仅链接链接而没有参数高亮显示时。单击时如何突出显示其他链接? 我生成菜单的代码是:
dmstr\widgets\Menu::widget(
[
'options' => ['class' => 'sidebar-menu'],
'items' => [
[
'label' => 'Leave',
'icon' => 'share',
'url' => '#',
'visible'=> (General::ifUserPermitted('leaveCreate')||General::ifUserPermitted('leaveDeleteOwn')||General::ifUserPermitted('leaveReject')||General::ifUserPermitted('leaveSeeAllLeave')||General::ifUserPermitted('leaveTeamLeaveApproval')||General::ifUserPermitted('leaveViewOwn')),
'items' => [
[
'label' => 'Create',
'icon' => 'creative-commons',
'url' => ['/leaveparent/create'],
'visible'=> General::ifUserPermitted('leaveCreate')
],
[
'label' => 'View',
'icon' => 'address-book-o ',
'url' => ['/leaveparent/index', 'LeaveparentSearch[userName]' => common\models\Person::getFirstNameByUserId(Yii::$app->user->id)],
'visible'=> General::ifUserPermitted('leaveViewOwn')
],
[
'label' => 'View All',
'icon' => 'address-book-o ',
'url' => ['/leaveparent/index'],
'visible'=> General::ifUserPermitted('leaveSeeAllLeave')
],
[
'label' => 'Pending for approval',
'icon' => 'check',
'url' => ['/leaveparent/index', 'LeaveparentSearch[report_to_user_id]' => Yii::$app->user->id, 'LeaveparentSearch[status]' => 'Pending'],
'visible'=> (General::ifUserPermitted('leaveTeamLeaveApproval')||General::ifUserPermitted('leaveReject'))
],
],
],
]
]
)
答案 0 :(得分:0)
在要设置为活动的项目配置中添加active
配置密钥。当条件满足时,它应该返回true
。类似的东西:
'active' => Yii::$app->controller->id === 'leaveparent'
&& Yii::$app->controller->action->id === 'index',