如何在 child 处于活动状态时使 parent 状态处于活动状态(即添加活动 class)
how to make parent state active(i.e., add active class) when child is active
我目前有一些路线
<ul class="">
<li class="">
<a [routerLinkActive]="['active']"
[routerLink]="['/user/traffic/graph']" class="">
Traffic
</a>
</li>
</ul>
路线
const trafficRoutes: Routes = [
{
path: '', component: TrafficComponent, children: [
{ path: '', redirectTo: 'graph' },
{ path: 'graph', component: GraphComponent},
{ path: 'alerts', component: AlertComponent}
]
},
];
当我导航到 GraphComponent
时,路由器 Link 被激活,但当我导航到 AlertComponent
时,它没有激活。
我想激活路由器 link。即使我导航到 AlertComponent
如何实现?
我的错。菜鸟的错误。更改为模块 URL 修复一切
<ul class="">
<li class="">
<a [routerLinkActive]="['active']"
[routerLink]="['/user/traffic/']" class="">
Traffic
</a>
</li>
</ul>
我目前有一些路线
<ul class="">
<li class="">
<a [routerLinkActive]="['active']"
[routerLink]="['/user/traffic/graph']" class="">
Traffic
</a>
</li>
</ul>
路线
const trafficRoutes: Routes = [
{
path: '', component: TrafficComponent, children: [
{ path: '', redirectTo: 'graph' },
{ path: 'graph', component: GraphComponent},
{ path: 'alerts', component: AlertComponent}
]
},
];
当我导航到 GraphComponent
时,路由器 Link 被激活,但当我导航到 AlertComponent
时,它没有激活。
我想激活路由器 link。即使我导航到 AlertComponent
如何实现?
我的错。菜鸟的错误。更改为模块 URL 修复一切
<ul class="">
<li class="">
<a [routerLinkActive]="['active']"
[routerLink]="['/user/traffic/']" class="">
Traffic
</a>
</li>
</ul>