我遇到一种情况,我想在第三级操作中访问动作有效负载。 我可以在可操作的运算符中进行这种操作,但是如何使用可管道运算符进行相同操作?
这是我的代码,
@Effect()
onTrySignin = this.actions$.pipe(
ofType(AuthActions.TRY_SIGNIN),
map((action: AuthActions.TrySignin) => {
return action.payload;
}),
switchMap(action => {
return this.httpService
.postRequest('UserAccounts/Login', action.credentials);
}), catchError((error: HttpErrorResponse) => {
return Observable.of(new AuthActions.FailedAuth(error));
}),
mergeMap((response: any) => {
// how to access action payload here?
})
);
答案 0 :(得分:1)
您可以使用map()
沿这样的可观察链传递数据:
// both foo and bar will be available on next()
from(AsyncFooData()).pipe(
concatMap(foo => AsyncBarData().pipe(
map(bar => ({foo, bar})
)),
tap(val => console.log(val), // chain more operators here...
).subscribe(({foo, bar}) => {
// do stuff with foo and bar
})
FWIW,我从this question那里得到了这个答案,我在其中发布了一个类似的答案。
答案 1 :(得分:0)
好吧,它是pipe
内的pipe
@Effect()
onTrySignin = this.actions$.pipe(
ofType(AuthActions.TRY_SIGNIN),
map((action: AuthActions.TrySignin) => {
return action.payload;
}),
switchMap(actionPayload => {
return this.httpService.postRequest('UserAccounts/Login', actionPayload.credentials).pipe(
mergeMap((response: HttpResponse<IApiResponder<string>>) => {
switch (response.status) {
case 200:
if (actionPayload.returnUrl) {
this.router.navigate([actionPayload.returnUrl]);
} else {
this.router.navigate(['/dbapp']);
}
return Observable.concat(
Observable.of(new AuthActions.GenerateAntiforgeryToken()),
Observable.of(new AuthActions.Signin(this.authService.getUserData())),
);
}
}),
catchError(e => {
return Observable.of(new AuthActions.FailedAuth(e));
}),
);
}),
);