是否有一种方法可以完成这种方法而又不复制children
属性的代码?
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},
]
}
]
}
];
我对member-profile/:mcid
和member-profile/:mcid/:hcid
的路线有重复的代码,我不太喜欢。我试图创建一个可以创建这两个对象的函数,但是Angular抱怨我无法在模板中使用方法。
如果我将路线用作member-profile/:mcid/:hcid?
,则也不起作用。
任何更好的方法?
// 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');