[英]Angular router with optional parameter
有沒有辦法在不復制children
財產代碼的情況下完成這種方法?
const routes: Routes = [
{
path: '', component: SmartSearchComponent, canActivate: [AuthGuard],
canActivateChild: [AuthGuard],
children: [
{path: '', redirectTo: 'home', pathMatch: 'full'},
{path: 'home', component: HomeComponent},
{
path: 'member-profile/:mcid',
component: MemberProfileComponent,
children: [
{path: '', redirectTo: 'member-info', pathMatch: 'full'},
{path: 'member-info', pathMatch: 'full', component: MemberInfoComponent},
{path: 'id-cards', component: IdCardsComponent},
{path: 'register-family-members', component: RegisteredFamilyMembersComponent},
{path: 'associate-caregivers', component: AssociateCaregiversComponent},
{path: 'member-preferences', component: MemberPreferencesComponent},
{path: 'two-fa-info', component: TwofaInfoComponent},
{path: 'coverage/:hcid', component: CoverageComponent},
]
},
{
path: 'member-profile/:mcid/:hcid',
component: MemberProfileComponent,
children: [
{path: '', redirectTo: 'member-info', pathMatch: 'full'},
{path: 'member-info', pathMatch: 'full', component: MemberInfoComponent},
{path: 'id-cards', component: IdCardsComponent},
{path: 'register-family-members', component: RegisteredFamilyMembersComponent},
{path: 'associate-caregivers', component: AssociateCaregiversComponent},
{path: 'member-preferences', component: MemberPreferencesComponent},
{path: 'two-fa-info', component: TwofaInfoComponent},
{path: 'coverage/:hcid', component: CoverageComponent},
]
}
]
}
];
我不太喜歡路由member-profile/:mcid
和member-profile/:mcid/:hcid
hcid 的代碼重復。 我試圖有一個 function 來創建這兩個對象,但 Angular 抱怨我不能在模板中有方法。
如果我將路由用作member-profile/:mcid/:hcid?
它也不起作用。
有什么更好的方法嗎?
// Declaration
const appRoutes: Routes = [{ path: employee, component: abcComp }]
// You dont need to set Query parameter in routing
// Implementation in Ts
this.route.navigate(['employee'], { queryParams: { name: 'a' } })
//queryParamsHandling='Merge' or 'retain' these options is also used to retain the parameters or to merge them
// Catching in Ts
constructor(private route : ActivateRoute){
}
let name = this.route.snapshot.queryParamMap.get('name');
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.