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我试图让一个组件具有一个由page-router-outlet
访问的ActionBar和子组件。为什么此路由不通过`page-router-outlet`路由?
我父组件的HTML如下:
<ActionBar class="action-bar">
<NavigationButton ios:visibility="collapsed" icon="res://menu" (tap)="onDrawerButtonTap()"></NavigationButton>
<ActionItem icon="res://navigation/menu" android:visibility="collapsed" (tap)="onDrawerButtonTap()"
ios.position="left">
</ActionItem>
<Label class="action-bar-title" text="Example Text"></Label>
</ActionBar>
<RadSideDrawer #drawer showOverNavigation="true" [drawerTransition]="sideDrawerTransition">
<StackLayout tkDrawerContent>
<MyDrawer [selectedPage]="'Settings'"></MyDrawer>
</StackLayout>
<StackLayout class="page page-content" tkMainContent>
<page-router-outlet></page-router-outlet>
</StackLayout>
</RadSideDrawer>
我routing-module
父组件的样子:
import {NgModule} from '@angular/core';
import {Routes} from '@angular/router';
import {NativeScriptRouterModule} from 'nativescript-angular/router';
import {HomeComponent} from './home.component';
import {DashboardComponent} from '../dashboard/dashboard.component';
import {Dashboard2Component} from '../dashboard2/dashboard.component';
const routes: Routes = [
{
path: '',
component: HomeComponent,
children: [
{path: '', component: DashboardComponent},
{path: 'projects', component: Dashboard2Component},
]
}
];
@NgModule({
imports: [NativeScriptRouterModule.forChild(routes)],
exports: [NativeScriptRouterModule]
})
export class HomeRoutingModule {
}
export const routedComponents = [
HomeComponent
];
然而,当我的路线/projects
,它带给我的带新操作栏的新页面
我的理解是,这是因为page-router-outlet
创建了一个新页面,据说我希望能够从一个子组件导回到以前的子组件。 {N}可能吗?
您可以尝试使用''而不是''...第一个不会创建一个新页面,只能替换当前内容中的内容... –
@AndersonIvanWitzke但是,如果我这样做,它不会允许我按后退按钮以访问上一页 – Crutchcorn