如果在 2 个延迟加载模块中使用了 Angular 服务，则为providedIn 设置什么

Question

Since Angular 6 recommended way is to use providedIn property of Injectable .

But what if need to use service in 2 lazy loaded modules?
Does creating shared module and then import shared module in both lazy loaded modules consider still the best practice?

Something like this(following this link to avoid circular reference warning):

//Shared module
@NgModule({})
export class SharedModule { }

//Shared service
@Injectable({
  providedIn: SharedModule
})
export class SharedService { }


//Lazy loaded modules A and B
@NgModule({
  imports: [SharedModule]
})
export class LazyModuleA { }

@NgModule({
  imports: [SharedModule]
})
export class LazyModuleB { }

Edit based on first answer: There will be quite (40-80) services.

Answer 1

I don't thing it is a good idea.Because you inject a service in an injector for situations that may or may not happen .When you inject this service in ShareModule, this service is alive until your app is alive.Let's imagine a situation that you've got lots of these scenarios, you inject a lot of services in injector that is memory consuming. I think it's better to inject this service at level of lazyLoad modules.

Answer 2

This is what I have observed with services in Angular 6(and above)

You can safely use "providedIn: 'root'"

The only catch is please do not add it in imports in the app.module (or whatever entry module you have). If you add it, it will load immediately.

Else, if you use "providedIn: 'root'" and it is getting used in any number of modules, it will not load until any of the module that uses that service loads.

Once loaded, it shares the same reference across modules whoever is using it.

Example description:

Service A(provided in root)

app.module

lazy-module-1(does not use service A)

lazy-module-2(uses service A)

lazy-module-3(does not use A)

lazy-module-4(uses A)

service A will load when either lazy-module-2 or lazy-module-4 is loaded. Once loaded, it stays in the app

Do let me know if you want a demo app on the same.