Question

我有一个自定义异常处理程序，如果发生任何异常（仅尝试它），它应该将用户带到自定义错误页面。

我正在尝试使用Injector获取路由器的实例。这样做的原因，我相信注入器将给现有路由器实例并使用它我将能够路由用户。

为什么这不起作用或如何实现这一点的任何想法？

谢谢你：）

@Injectable()
export class AppExceptionHandler extends ExceptionHandler{

constructor(){
    super(null, null);
}

call(exception:any, stackTrace?:any, reason?:string):void {
    console.log('call...')


    var providers = Injector.resolve([ROUTER_PROVIDERS]);
    var injector = Injector.fromResolvedProviders(providers);

    // this is causing issue, not sure it is the correct way
    let router : Router = injector.get(Router);

    // not executed
    console.log(router)

    // not executed 
    console.log('done...')
    router.navigate(["CustomErrorPage"]);
    }

}

回答 - 在2.0.0-beta.17中测试过感谢Druxtan

1. Created a file app.injector.ts inside the app folder (app/app.injector.ts)

let appInjectorRef;

export const appInjector = (injector?) => {
    if (!injector) {
        return appInjectorRef;
    }

    appInjectorRef = injector;

    return appInjectorRef;
};

2. Added to the bootstrap in the main.ts

bootstrap(AppComponent,[ROUTER_PROVIDERS, HTTP_PROVIDERS,provide(ExceptionHandler,{useClass : AppExceptionHandler})])
    .then((appRef) => appInjector(appRef.injector));

3. In the AppExceptionHandler, retrieved the Router instance as shown below

export class AppExceptionHandler {

    call(exception:any, stackTrace?:any, reason?:string):void {

        let injectorApp = appInjector();
        let router = injectorApp.get(Router);
        let localStorageService = injectorApp.get(LocalStorageService);

        if(exception.message === '-1'){
            localStorageService.clear();
            router.navigate(["Login"]);
        }
    }

}

Answer 1

我会以这种方式实现你的功能，因为这个类发生在依赖注入：

@Injectable()
export class AppExceptionHandler extends ExceptionHandler {
  constructor(private router:Router) {
    super(null, null);
  }

  call(exception:any, stackTrace?:any, reason?:string):void {
    console.log('call...')

    this.router.navigate(['CustomErrorPage']);
  }
}

以这种方式注册你的句柄：

bootstrap(MyApp, [
  provide(ExceptionHandler, {useClass: AppExceptionHandler})
]);

Angular 2自定义错误处理和路由器

1 个答案: