In Angular 6 how make case insensitive url pattern?

Question

In my case I want to support same url in case insensitive manner.

Example: it should support all url

localhost:1029/documentation
localhost:1029/DOCU         


        

Answer 1

This will work, configure the route to NotFoundComponent with wild character like below

{path:'**',component:NotFoundComponent}

then, in the NotFoundComponent.ts file, add the below lines inside ngOnInit()

if(this.route.snapshot.url[0].path.toLowerCase()!==this.route.snapshot.url[0].path)
      this.router.navigate([this.route.snapshot.url[0].path.toLowerCase()]);

You have to import ActivatedRoute,Router from '@angular/router' and inject in constructor like below

constructor(private route:ActivatedRoute,
    private router:Router) { }

Here if condition in ngOnInit() makes sure that it won't route or navigate infinitely

Answer 2

You need a UrlSerializer as follow:

import { DefaultUrlSerializer, UrlTree } from '@angular/router';


 export class LowerCaseUrlSerializer extends DefaultUrlSerializer {
   parse(url: string): UrlTree {
      return super.parse(url.toLowerCase());
  }
}

And then added it as a provider in the app.module.ts

providers: [
 {
   provide: UrlSerializer,
   useClass: LowerCaseUrlSerializer
}
]

Answer 3

You should add this provide statement to the app.module.ts

import { DefaultUrlSerializer, UrlTree } from '@angular/router';

export class LowerCaseUrlSerializer extends DefaultUrlSerializer {
    parse(url: string): UrlTree {
        // Optional Step: Do some stuff with the url if needed.

        // If you lower it in the optional step 
        // you don't need to use "toLowerCase" 
        // when you pass it down to the next function
        return super.parse(url.toLowerCase()); 
    }
}

And

@NgModule({
    imports: [
      ...
    ],
    declarations: [AppComponent],
    providers: [
        {
            provide: UrlSerializer,
            useClass: LowerCaseUrlSerializer
        }
    ],
    bootstrap: [AppComponent]
})