[英]Organizing Angular feature modules and their sub-modules with routing
我正在尝试遵循Angular中的功能模块方法。
我有一个带有管理模块的应用程序。 子模块(即用户管理,项目管理)内将包含子模块。
下面是我想出的结构:
app.component.css
app.component.html
app.component.ts
app.module.ts
app-routing.module.ts
admin/
admin.component.css
admin.component.html
admin.component.ts
admin.module.ts
admin-routing.module.ts
user-management/
user-management.component.css
user-management.component.html
user-management.component.ts
user-management.module.ts
user-management-routing.module.ts
user-list/
user-list.component.css
user-list.component.html
user-list.component.ts
user-detail/
user-detail.component.css
user-detail.component.html
user-detail.component.ts
project-management/
project-management.component.css
project-management.component.html
project-management.component.ts
project-management.module.ts
project-management-routing.module.ts
project-list/
project-list.component.css
project-list.component.html
project-list.component.ts
project-detail/
project-detail.component.css
project-detail.component.html
project-detail.component.ts
如果我错了,请指正我,但是我相信上面的结构就像它的教科书一样。 我真的很努力地使路由使用这种方法。
请在下面查看我的路线:
const appRoutes: Routes = [
{ path: 'admin', loadChildren: './admin/admin.module#AdminModule', canLoad:
[AuthGuard] },
{ path: '', redirectTo: '/login', pathMatch: 'full' },
{ path: '**', component: PageNotFoundComponent }
];
const adminRoutes: Routes = [
{
path: '',
component: AdminComponent,
canActivate: [AuthGuard],
children: [
{
path: '',
canActivateChild: [AuthGuard],
children: [
{ path: 'users', component: UserManagementComponent },
{ path: 'projects', component: ProjectManagementComponent }
]
}
]
}
];
const manageUsers: Routes = [
{path: 'users', component: UserListComponent},
{path: 'users/edit/:id', component: UserDetailComponent}
];
const manageProjects: Routes = [
{path: 'projects', component: ProjectListComponent},
{path: 'projects/edit/:id', component: ProjectDetailComponent}
];
我希望导航的方式如下:
/ admin(除admin.component.html布局外不显示任何内容-某天将显示一个仪表板组件)
/ admin / users(显示用户列表)
/ admin / users / edit / 1(显示用户的详细信息)
我认为我的主要难题是我拥有一个用户管理组件,默认情况下,真正应该显示在其上的是用户列表组件。 admin.component.html中有一个路由器出口,而user-management.component.html中有一个路由器出口。 像这样的事情是否需要一个双重出口?还是我应该直接在user-management-component.html页面上显示而不是路由器出口?
这是一个与您建议的模块组件层次结构直接对应的路由结构:
const appRoutes: Routes = [
{
path: 'admin',
component: AdminComponent, // has <router-outlet>
children: [
{
path: 'users',
component: UserManagementComponent, // has <router-outlet>
children: [
{
path: '',
component: UserListComponent
},
{
path: 'edit/:id',
component: UserDetailComponent
}
]
},
{
path: 'projects',
component: ProjectManagementComponent, // has <router-outlet>
children: [
{
path: '',
component: ProjectListComponent
},
{
path: 'edit/:id',
component: ProjectDetailComponent
}
]
}
]
},
{ path: '', redirectTo: '/admin' },
{ path: '**', component: PageNotFoundComponent }
];
这将为您提供一个主要的“管理”页面,该页面除了模板中的HTML之外(如果包含<router-outlet>
之外的任何内容),什么都不显示。 导航到/admin/users
,它将在UserManaementComponent
内部显示UserListComponent
(考虑到UserManagementComponent
具有<router-outlet>
)。 如果单击指向/admin/users/edit/:id
(其中:id
是一些ID),则UserListComponent
将与UserManagementComponent
内部的UserDetailComponent
交换。
同样对于“项目”方面。
如果要细分为功能模块,可以这样做:
app-routing/module.ts
// From app-routing.module.ts
const appRoutes: Routes = [
{
path: 'admin',
loadChildren: './admin/admin.module#AdminModule',
canLoad: [AuthGuard]
},
{ path: '', redirectTo: '/login', pathMatch: 'full' },
{ path: '**', component: PageNotFoundComponent }
];
/admin/admin-routing.module.ts
// From /admin/admin-routing.module.ts
const ADMIN_ROUTES: Routes = [
{
path: '',
component: AdminComponent,
canActivate: [AuthGuard],
children: [
{
path: '',
canActivateChild: [AuthGuard],
children: [
{ path: 'users', loadChildren: './admin/user-management/user-management.module#UserManagementModule' },
{ path: 'projects', loadChildren: './admin/project-management/project-management.module#ProjectManagementModule' }
]
}
]
}
];
/admin/user-management/user-management-routing.module.ts
// From /admin/user-management/user-management-routing.module.ts
const USER_MANAGEMENT_ROUTES: Routes = [
{path: '', component: UserListComponent},
{path: 'edit/:id', component: UserDetailComponent}
];
/admin/project-management/project-management-routing.module.ts
// From /admin/project-management/project-management-routing.module.ts
const PROJECT_MANAGEMENT_ROUTES: Routes = [
{path: '', component: ProjectListComponent},
{path: 'edit/:id', component: ProjectDetailComponent}
];
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