Angular2基于搜索结果路由到页面

Question

如果搜索结果只有一个项目（如果reportId = 1），我想导航到不同的页面，用户可以保存一次点击。这是我的解析器代码。

export class SearchResultsResolver implements Resolve<Report[]> {
        constructor(private searchService: SearchService, private router: Router) {}

    resolve(route: ActivatedRouteSnapshot,
         state: RouterStateSnapshot): Observable<Report[]> {
        let reportId = route.params['reportId'];

        return this.searchService.getReports(reportId);

        console.log('reportid', reportId);
    }
}

Answer 1

您只需使用flatMap，Observer.if和router.navigate的组合即可完成，如下面的代码段所示

resolve(route: ActivatedRouteSnapshot, state: RouterStateSnapshot): Observable<Array<Report>> {
  let reportId = route.params['reportId'];

  return this.searchService.getReports(reportId)
 .flatMap(reports => Observable.if(() => reports && reports.length > 1, 
      Observable.of(reports),
      this._router.navigate(['/otherRoute']))
  );