我的app结构如下
.
├── photos
├── posts
├── users
│ ├── detail
│ │ ├── address
│ │ ├── family
│ │ ├── information
│ │ └── phones
│ ├── friends
│ └── profile
└── videos
要创建嵌套路由,我更喜欢自己级别的相同级别的路由。
例如:
第一级路由进入根路由。第二级路由转到用户路由,第三级转到详细路由。
所以根路由看起来像这样,
const appRoutes: Routes = [
{
path: '',
component: HomeComponent
},
{
path: 'photos',
component: PhotosComponent
},
{
path: 'posts',
component: PostsComponent
},
{
path: 'users',
component: UserListComponent
},
{
path: 'user/:username',
component: UserComponent
},
{
path: 'videos',
component: VideosComponent
}
]
export const AppRoutes = RouterModule.forRoot(appRoutes);
// and the for the user routes,
const userRoutes: Routes = [
{
path: 'detail',
component: DetailComponent
},
{
path: 'friends',
component: FriendsComponent
},
{
path: 'profile',
component: ProfileComponent
}
]
export const UserRoutes = RouterModule.forChild(userRoutes);
// and for the detail routes,
const detailRoutes: Routes = [
{
path: 'address',
component: AddressComponent
},
{
path: 'family',
component: FamilyComponent
},
{
path: 'info',
component: InformationComponent
},
{
path: 'phones',
component: PhonesComponent
}
]
export const DetailRoutes = RouterModule.forChild(detailRoutes);
如何混淆给定的路线?
我希望路线为/users/:username/detail/info
,但无法弄清楚如何将它们联系起来。
答案 0 :(得分:0)
您必须使用children
属性设置每条路线的子项。如果你想将它们保存在单独的文件中,我想你可以导出Routes
的const数组(而不是RouterModule
)并导入它们,以便通过这样的方式使用它们:
<强> user.routes 强>
export const userRoutes : Routes = [
{
path: 'detail',
component: DetailComponent
},
{
path: 'friends',
component: FriendsComponent
},
{
path: 'profile',
component: ProfileComponent
}
]
<强> app.routes 强>
import {userRoutes} from './user/user.routes'
const appRoutes: Routes = [
{
path: '',
component: HomeComponent
},
{
path: 'photos',
component: PhotosComponent
},
{
path: 'posts',
component: PostsComponent
},
{
path: 'users',
component: UserListComponent
},
{
path: 'user/:username',
component: UserComponent,
children: [userRoutes[0]]
},
{
path: 'videos',
component: VideosComponent
}
]
export const AppRoutes = RouterModule.forRoot(appRoutes);