Defining relative routing in Angular









up vote
0
down vote

favorite












Update



I changed forRoot with forChild according to the answers.



So, basically I have to issues, let this be a submodule,



@NgModule(
imports: [
CommonModule,
ARoutingModule,
BModule
],
declarations: [AppsComponent, ...],
exports: [
AppsComponent, ...
]
)
export class AModule


and



@NgModule(
imports: [
CommonModule,
BRoutingModule
],
declarations: [AppsComponent, ...],
exports: [
AppsComponent, ...
]
)
export class BModule


So AModule is being imported by the root module and both modules, AModule and BModule shall define their own routes, something like



// ./A/A-Routing.module.ts
const routes: Routes = [

path: 'A',//<-- this shall route to www.site.com/A
component: AComponent

];

@NgModule(
imports: [RouterModule.forChild(routes)],
exports: [RouterModule],
declarations:
)
export class ARoutingModule


and in SubSub I have



// ./A/B/B-Routing.module.ts
const routes: Routes = [

path: 'B', //<-- this shall route to www.site.com/A/B
component: BComponent,

]

];

@NgModule(
imports: [RouterModule.forChild(routes)],
exports: [RouterModule],
declarations:
)
export class BRoutingModule


Is this possible? When using a path printing, I can get the sub-routes, but not the subsub routes (i.e. B). And can I define the routes in B without knowing the path before that? so define B without knowing A



www.site.com/something/else


define routes for else without knowing something?



Here is a stackblitz example, main works, but the Subs don't...










share|improve this question























  • RouterModule.forRoot() is used for routing only in the root module of the application. For using routes in other modules, use RouterModule.forChild()
    – Sachin Gupta
    Nov 10 at 16:56














up vote
0
down vote

favorite












Update



I changed forRoot with forChild according to the answers.



So, basically I have to issues, let this be a submodule,



@NgModule(
imports: [
CommonModule,
ARoutingModule,
BModule
],
declarations: [AppsComponent, ...],
exports: [
AppsComponent, ...
]
)
export class AModule


and



@NgModule(
imports: [
CommonModule,
BRoutingModule
],
declarations: [AppsComponent, ...],
exports: [
AppsComponent, ...
]
)
export class BModule


So AModule is being imported by the root module and both modules, AModule and BModule shall define their own routes, something like



// ./A/A-Routing.module.ts
const routes: Routes = [

path: 'A',//<-- this shall route to www.site.com/A
component: AComponent

];

@NgModule(
imports: [RouterModule.forChild(routes)],
exports: [RouterModule],
declarations:
)
export class ARoutingModule


and in SubSub I have



// ./A/B/B-Routing.module.ts
const routes: Routes = [

path: 'B', //<-- this shall route to www.site.com/A/B
component: BComponent,

]

];

@NgModule(
imports: [RouterModule.forChild(routes)],
exports: [RouterModule],
declarations:
)
export class BRoutingModule


Is this possible? When using a path printing, I can get the sub-routes, but not the subsub routes (i.e. B). And can I define the routes in B without knowing the path before that? so define B without knowing A



www.site.com/something/else


define routes for else without knowing something?



Here is a stackblitz example, main works, but the Subs don't...










share|improve this question























  • RouterModule.forRoot() is used for routing only in the root module of the application. For using routes in other modules, use RouterModule.forChild()
    – Sachin Gupta
    Nov 10 at 16:56












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Update



I changed forRoot with forChild according to the answers.



So, basically I have to issues, let this be a submodule,



@NgModule(
imports: [
CommonModule,
ARoutingModule,
BModule
],
declarations: [AppsComponent, ...],
exports: [
AppsComponent, ...
]
)
export class AModule


and



@NgModule(
imports: [
CommonModule,
BRoutingModule
],
declarations: [AppsComponent, ...],
exports: [
AppsComponent, ...
]
)
export class BModule


So AModule is being imported by the root module and both modules, AModule and BModule shall define their own routes, something like



// ./A/A-Routing.module.ts
const routes: Routes = [

path: 'A',//<-- this shall route to www.site.com/A
component: AComponent

];

@NgModule(
imports: [RouterModule.forChild(routes)],
exports: [RouterModule],
declarations:
)
export class ARoutingModule


and in SubSub I have



// ./A/B/B-Routing.module.ts
const routes: Routes = [

path: 'B', //<-- this shall route to www.site.com/A/B
component: BComponent,

]

];

@NgModule(
imports: [RouterModule.forChild(routes)],
exports: [RouterModule],
declarations:
)
export class BRoutingModule


Is this possible? When using a path printing, I can get the sub-routes, but not the subsub routes (i.e. B). And can I define the routes in B without knowing the path before that? so define B without knowing A



www.site.com/something/else


define routes for else without knowing something?



Here is a stackblitz example, main works, but the Subs don't...










share|improve this question















Update



I changed forRoot with forChild according to the answers.



So, basically I have to issues, let this be a submodule,



@NgModule(
imports: [
CommonModule,
ARoutingModule,
BModule
],
declarations: [AppsComponent, ...],
exports: [
AppsComponent, ...
]
)
export class AModule


and



@NgModule(
imports: [
CommonModule,
BRoutingModule
],
declarations: [AppsComponent, ...],
exports: [
AppsComponent, ...
]
)
export class BModule


So AModule is being imported by the root module and both modules, AModule and BModule shall define their own routes, something like



// ./A/A-Routing.module.ts
const routes: Routes = [

path: 'A',//<-- this shall route to www.site.com/A
component: AComponent

];

@NgModule(
imports: [RouterModule.forChild(routes)],
exports: [RouterModule],
declarations:
)
export class ARoutingModule


and in SubSub I have



// ./A/B/B-Routing.module.ts
const routes: Routes = [

path: 'B', //<-- this shall route to www.site.com/A/B
component: BComponent,

]

];

@NgModule(
imports: [RouterModule.forChild(routes)],
exports: [RouterModule],
declarations:
)
export class BRoutingModule


Is this possible? When using a path printing, I can get the sub-routes, but not the subsub routes (i.e. B). And can I define the routes in B without knowing the path before that? so define B without knowing A



www.site.com/something/else


define routes for else without knowing something?



Here is a stackblitz example, main works, but the Subs don't...







angular typescript routes






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 10 at 17:52

























asked Nov 10 at 16:26









rst

1,536728




1,536728











  • RouterModule.forRoot() is used for routing only in the root module of the application. For using routes in other modules, use RouterModule.forChild()
    – Sachin Gupta
    Nov 10 at 16:56
















  • RouterModule.forRoot() is used for routing only in the root module of the application. For using routes in other modules, use RouterModule.forChild()
    – Sachin Gupta
    Nov 10 at 16:56















RouterModule.forRoot() is used for routing only in the root module of the application. For using routes in other modules, use RouterModule.forChild()
– Sachin Gupta
Nov 10 at 16:56




RouterModule.forRoot() is used for routing only in the root module of the application. For using routes in other modules, use RouterModule.forChild()
– Sachin Gupta
Nov 10 at 16:56












2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










There are a few things that you'll have to fix to make it work properly.



  1. You'll have to add an opening route for your SubModule which will be a Lazy Loaded Module. You'll do this in your main-routing.module.ts file


const routes: Routes = [

path: 'main',
component: MainComponent,
children: [
path: 'sub',
loadChildren: './sub/sub.module#SubModule'
]

];


  1. You'll then create a module named SubModule and the SubRoutingModule will act as a routing module for this SubModule. You'll declare your Sub1Component and Sub2Component here:


import Sub1Component, Sub2Component from './sub.component';
import NgModule from '@angular/core';
import Routes, RouterModule from '@angular/router';

import SubRoutingModule from './sub-routing.module';

@NgModule(
declarations: [Sub1Component, Sub2Component],
imports: [SubRoutingModule]
)
export class SubModule


So now your routes will change to :



<a routerLink="/main/sub/sub1">Sub1</a> | 
<a routerLink="/main/sub/sub2">Sub2</a>


Here's an Updated StackBlitz for your ref.






share|improve this answer






















  • I need to have multiple routing files at different locations.
    – rst
    Nov 10 at 17:18










  • There's a serious architectural issue with your Current Implementation. Can you please share a more concrete example of your modules and probably a minimal stackblitz sample replicating your issue?
    – SiddAjmera
    Nov 10 at 17:19










  • Stackblitz example in the op.
    – rst
    Nov 10 at 17:53










  • @rst, please go through my updated answer. Hopefully this should help you achieve what you're trying to.
    – SiddAjmera
    Nov 10 at 18:51

















up vote
1
down vote













Change forRoot to forChild in SubSubRoutingModule



// ./Sub/subsub/SubSub-Routing.module.ts
const routes: Routes = [

path: 'subsub', //<-- this shall route to www.site.com/sub/subsub
component: SubSubComponent,
]

];

@NgModule(
imports: [RouterModule.forChild(routes)], //<-- change here
exports: [RouterModule],
declarations:
)
export class SubSubRoutingModule





share|improve this answer




















  • That works, but SubSub (I changed it in the OP to A and B, thought that was more clearer) get routet to the root, i.e. /subsub instead of /Sub/subsub
    – rst
    Nov 10 at 17:15










  • You should implement with Lazy Module. Look at this demo for reference stackblitz.com/edit/angular-hcerdo
    – Sunil Singh
    Nov 10 at 17:46










  • I have added a stackblitz example
    – rst
    Nov 10 at 17:52






  • 1




    Here is the modified version of your demo stackblitz.com/edit/angular-3mjttd
    – Sunil Singh
    Nov 10 at 18:04










  • That's exactly what I was Looking for! Excellent
    – rst
    Nov 10 at 19:16










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










There are a few things that you'll have to fix to make it work properly.



  1. You'll have to add an opening route for your SubModule which will be a Lazy Loaded Module. You'll do this in your main-routing.module.ts file


const routes: Routes = [

path: 'main',
component: MainComponent,
children: [
path: 'sub',
loadChildren: './sub/sub.module#SubModule'
]

];


  1. You'll then create a module named SubModule and the SubRoutingModule will act as a routing module for this SubModule. You'll declare your Sub1Component and Sub2Component here:


import Sub1Component, Sub2Component from './sub.component';
import NgModule from '@angular/core';
import Routes, RouterModule from '@angular/router';

import SubRoutingModule from './sub-routing.module';

@NgModule(
declarations: [Sub1Component, Sub2Component],
imports: [SubRoutingModule]
)
export class SubModule


So now your routes will change to :



<a routerLink="/main/sub/sub1">Sub1</a> | 
<a routerLink="/main/sub/sub2">Sub2</a>


Here's an Updated StackBlitz for your ref.






share|improve this answer






















  • I need to have multiple routing files at different locations.
    – rst
    Nov 10 at 17:18










  • There's a serious architectural issue with your Current Implementation. Can you please share a more concrete example of your modules and probably a minimal stackblitz sample replicating your issue?
    – SiddAjmera
    Nov 10 at 17:19










  • Stackblitz example in the op.
    – rst
    Nov 10 at 17:53










  • @rst, please go through my updated answer. Hopefully this should help you achieve what you're trying to.
    – SiddAjmera
    Nov 10 at 18:51














up vote
1
down vote



accepted










There are a few things that you'll have to fix to make it work properly.



  1. You'll have to add an opening route for your SubModule which will be a Lazy Loaded Module. You'll do this in your main-routing.module.ts file


const routes: Routes = [

path: 'main',
component: MainComponent,
children: [
path: 'sub',
loadChildren: './sub/sub.module#SubModule'
]

];


  1. You'll then create a module named SubModule and the SubRoutingModule will act as a routing module for this SubModule. You'll declare your Sub1Component and Sub2Component here:


import Sub1Component, Sub2Component from './sub.component';
import NgModule from '@angular/core';
import Routes, RouterModule from '@angular/router';

import SubRoutingModule from './sub-routing.module';

@NgModule(
declarations: [Sub1Component, Sub2Component],
imports: [SubRoutingModule]
)
export class SubModule


So now your routes will change to :



<a routerLink="/main/sub/sub1">Sub1</a> | 
<a routerLink="/main/sub/sub2">Sub2</a>


Here's an Updated StackBlitz for your ref.






share|improve this answer






















  • I need to have multiple routing files at different locations.
    – rst
    Nov 10 at 17:18










  • There's a serious architectural issue with your Current Implementation. Can you please share a more concrete example of your modules and probably a minimal stackblitz sample replicating your issue?
    – SiddAjmera
    Nov 10 at 17:19










  • Stackblitz example in the op.
    – rst
    Nov 10 at 17:53










  • @rst, please go through my updated answer. Hopefully this should help you achieve what you're trying to.
    – SiddAjmera
    Nov 10 at 18:51












up vote
1
down vote



accepted







up vote
1
down vote



accepted






There are a few things that you'll have to fix to make it work properly.



  1. You'll have to add an opening route for your SubModule which will be a Lazy Loaded Module. You'll do this in your main-routing.module.ts file


const routes: Routes = [

path: 'main',
component: MainComponent,
children: [
path: 'sub',
loadChildren: './sub/sub.module#SubModule'
]

];


  1. You'll then create a module named SubModule and the SubRoutingModule will act as a routing module for this SubModule. You'll declare your Sub1Component and Sub2Component here:


import Sub1Component, Sub2Component from './sub.component';
import NgModule from '@angular/core';
import Routes, RouterModule from '@angular/router';

import SubRoutingModule from './sub-routing.module';

@NgModule(
declarations: [Sub1Component, Sub2Component],
imports: [SubRoutingModule]
)
export class SubModule


So now your routes will change to :



<a routerLink="/main/sub/sub1">Sub1</a> | 
<a routerLink="/main/sub/sub2">Sub2</a>


Here's an Updated StackBlitz for your ref.






share|improve this answer














There are a few things that you'll have to fix to make it work properly.



  1. You'll have to add an opening route for your SubModule which will be a Lazy Loaded Module. You'll do this in your main-routing.module.ts file


const routes: Routes = [

path: 'main',
component: MainComponent,
children: [
path: 'sub',
loadChildren: './sub/sub.module#SubModule'
]

];


  1. You'll then create a module named SubModule and the SubRoutingModule will act as a routing module for this SubModule. You'll declare your Sub1Component and Sub2Component here:


import Sub1Component, Sub2Component from './sub.component';
import NgModule from '@angular/core';
import Routes, RouterModule from '@angular/router';

import SubRoutingModule from './sub-routing.module';

@NgModule(
declarations: [Sub1Component, Sub2Component],
imports: [SubRoutingModule]
)
export class SubModule


So now your routes will change to :



<a routerLink="/main/sub/sub1">Sub1</a> | 
<a routerLink="/main/sub/sub2">Sub2</a>


Here's an Updated StackBlitz for your ref.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 10 at 18:51

























answered Nov 10 at 16:29









SiddAjmera

12k21137




12k21137











  • I need to have multiple routing files at different locations.
    – rst
    Nov 10 at 17:18










  • There's a serious architectural issue with your Current Implementation. Can you please share a more concrete example of your modules and probably a minimal stackblitz sample replicating your issue?
    – SiddAjmera
    Nov 10 at 17:19










  • Stackblitz example in the op.
    – rst
    Nov 10 at 17:53










  • @rst, please go through my updated answer. Hopefully this should help you achieve what you're trying to.
    – SiddAjmera
    Nov 10 at 18:51
















  • I need to have multiple routing files at different locations.
    – rst
    Nov 10 at 17:18










  • There's a serious architectural issue with your Current Implementation. Can you please share a more concrete example of your modules and probably a minimal stackblitz sample replicating your issue?
    – SiddAjmera
    Nov 10 at 17:19










  • Stackblitz example in the op.
    – rst
    Nov 10 at 17:53










  • @rst, please go through my updated answer. Hopefully this should help you achieve what you're trying to.
    – SiddAjmera
    Nov 10 at 18:51















I need to have multiple routing files at different locations.
– rst
Nov 10 at 17:18




I need to have multiple routing files at different locations.
– rst
Nov 10 at 17:18












There's a serious architectural issue with your Current Implementation. Can you please share a more concrete example of your modules and probably a minimal stackblitz sample replicating your issue?
– SiddAjmera
Nov 10 at 17:19




There's a serious architectural issue with your Current Implementation. Can you please share a more concrete example of your modules and probably a minimal stackblitz sample replicating your issue?
– SiddAjmera
Nov 10 at 17:19












Stackblitz example in the op.
– rst
Nov 10 at 17:53




Stackblitz example in the op.
– rst
Nov 10 at 17:53












@rst, please go through my updated answer. Hopefully this should help you achieve what you're trying to.
– SiddAjmera
Nov 10 at 18:51




@rst, please go through my updated answer. Hopefully this should help you achieve what you're trying to.
– SiddAjmera
Nov 10 at 18:51












up vote
1
down vote













Change forRoot to forChild in SubSubRoutingModule



// ./Sub/subsub/SubSub-Routing.module.ts
const routes: Routes = [

path: 'subsub', //<-- this shall route to www.site.com/sub/subsub
component: SubSubComponent,
]

];

@NgModule(
imports: [RouterModule.forChild(routes)], //<-- change here
exports: [RouterModule],
declarations:
)
export class SubSubRoutingModule





share|improve this answer




















  • That works, but SubSub (I changed it in the OP to A and B, thought that was more clearer) get routet to the root, i.e. /subsub instead of /Sub/subsub
    – rst
    Nov 10 at 17:15










  • You should implement with Lazy Module. Look at this demo for reference stackblitz.com/edit/angular-hcerdo
    – Sunil Singh
    Nov 10 at 17:46










  • I have added a stackblitz example
    – rst
    Nov 10 at 17:52






  • 1




    Here is the modified version of your demo stackblitz.com/edit/angular-3mjttd
    – Sunil Singh
    Nov 10 at 18:04










  • That's exactly what I was Looking for! Excellent
    – rst
    Nov 10 at 19:16














up vote
1
down vote













Change forRoot to forChild in SubSubRoutingModule



// ./Sub/subsub/SubSub-Routing.module.ts
const routes: Routes = [

path: 'subsub', //<-- this shall route to www.site.com/sub/subsub
component: SubSubComponent,
]

];

@NgModule(
imports: [RouterModule.forChild(routes)], //<-- change here
exports: [RouterModule],
declarations:
)
export class SubSubRoutingModule





share|improve this answer




















  • That works, but SubSub (I changed it in the OP to A and B, thought that was more clearer) get routet to the root, i.e. /subsub instead of /Sub/subsub
    – rst
    Nov 10 at 17:15










  • You should implement with Lazy Module. Look at this demo for reference stackblitz.com/edit/angular-hcerdo
    – Sunil Singh
    Nov 10 at 17:46










  • I have added a stackblitz example
    – rst
    Nov 10 at 17:52






  • 1




    Here is the modified version of your demo stackblitz.com/edit/angular-3mjttd
    – Sunil Singh
    Nov 10 at 18:04










  • That's exactly what I was Looking for! Excellent
    – rst
    Nov 10 at 19:16












up vote
1
down vote










up vote
1
down vote









Change forRoot to forChild in SubSubRoutingModule



// ./Sub/subsub/SubSub-Routing.module.ts
const routes: Routes = [

path: 'subsub', //<-- this shall route to www.site.com/sub/subsub
component: SubSubComponent,
]

];

@NgModule(
imports: [RouterModule.forChild(routes)], //<-- change here
exports: [RouterModule],
declarations:
)
export class SubSubRoutingModule





share|improve this answer












Change forRoot to forChild in SubSubRoutingModule



// ./Sub/subsub/SubSub-Routing.module.ts
const routes: Routes = [

path: 'subsub', //<-- this shall route to www.site.com/sub/subsub
component: SubSubComponent,
]

];

@NgModule(
imports: [RouterModule.forChild(routes)], //<-- change here
exports: [RouterModule],
declarations:
)
export class SubSubRoutingModule






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 10 at 16:37









Sunil Singh

6,1121626




6,1121626











  • That works, but SubSub (I changed it in the OP to A and B, thought that was more clearer) get routet to the root, i.e. /subsub instead of /Sub/subsub
    – rst
    Nov 10 at 17:15










  • You should implement with Lazy Module. Look at this demo for reference stackblitz.com/edit/angular-hcerdo
    – Sunil Singh
    Nov 10 at 17:46










  • I have added a stackblitz example
    – rst
    Nov 10 at 17:52






  • 1




    Here is the modified version of your demo stackblitz.com/edit/angular-3mjttd
    – Sunil Singh
    Nov 10 at 18:04










  • That's exactly what I was Looking for! Excellent
    – rst
    Nov 10 at 19:16
















  • That works, but SubSub (I changed it in the OP to A and B, thought that was more clearer) get routet to the root, i.e. /subsub instead of /Sub/subsub
    – rst
    Nov 10 at 17:15










  • You should implement with Lazy Module. Look at this demo for reference stackblitz.com/edit/angular-hcerdo
    – Sunil Singh
    Nov 10 at 17:46










  • I have added a stackblitz example
    – rst
    Nov 10 at 17:52






  • 1




    Here is the modified version of your demo stackblitz.com/edit/angular-3mjttd
    – Sunil Singh
    Nov 10 at 18:04










  • That's exactly what I was Looking for! Excellent
    – rst
    Nov 10 at 19:16















That works, but SubSub (I changed it in the OP to A and B, thought that was more clearer) get routet to the root, i.e. /subsub instead of /Sub/subsub
– rst
Nov 10 at 17:15




That works, but SubSub (I changed it in the OP to A and B, thought that was more clearer) get routet to the root, i.e. /subsub instead of /Sub/subsub
– rst
Nov 10 at 17:15












You should implement with Lazy Module. Look at this demo for reference stackblitz.com/edit/angular-hcerdo
– Sunil Singh
Nov 10 at 17:46




You should implement with Lazy Module. Look at this demo for reference stackblitz.com/edit/angular-hcerdo
– Sunil Singh
Nov 10 at 17:46












I have added a stackblitz example
– rst
Nov 10 at 17:52




I have added a stackblitz example
– rst
Nov 10 at 17:52




1




1




Here is the modified version of your demo stackblitz.com/edit/angular-3mjttd
– Sunil Singh
Nov 10 at 18:04




Here is the modified version of your demo stackblitz.com/edit/angular-3mjttd
– Sunil Singh
Nov 10 at 18:04












That's exactly what I was Looking for! Excellent
– rst
Nov 10 at 19:16




That's exactly what I was Looking for! Excellent
– rst
Nov 10 at 19:16

















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