Question

在Rxjs中，想象一个流将获取一些数据，然后使用shareReplay来保存最新值以进行优化。

const value = fetchData().pipe(
  shareReplay(1)
)

但是，如果该值在这种情况下（仅在这种情况下）可以到期，应该重新获取。

const value = fetchData().pipe(
  shareReplay(1),
  switchMap((value) => isExpired(value) ? fetchData() : of(value))
)

这不太正确，因为在第一次重新获取后，该值不再共享，并且每次都会重新获取。

如何在Rxjs中表达“保持最新值”和“无效”的功能？

Answer 1

您可以执行以下操作：

const value$ = fetchData().pipe(
  expand(res => timer(res.validForMs).pipe(switchMap(() => fetchData()))),
  shareReplay(1)
);

通过这种方式，缓存失效发生在shareReplay之前，并且每个订阅者还将获得最新的值。

我已经做了一个简单的例子来演示：

//  just mocks
// ---------------------------
const fetchData = () =>
  of({
    token: uuid(),
    validForMs: 2000
  });
// ---------------------------

const value$ = fetchData().pipe(
  expand(res => timer(res.validForMs).pipe(switchMap(() => fetchData()))),
  shareReplay(1)
);

merge(
  value$.pipe(
    tap(res => console.log(`Subscriber 1 received token ${res.token}`))
  ),
  value$.pipe(
    tap(res => console.log(`Subscriber 2 received token ${res.token}`))
  ),
  value$.pipe(
    tap(res => console.log(`Subscriber 3 received token ${res.token}`))
  )
).subscribe();

输出为：

Subscriber 1 received token f7facb6f-2c16-4b0f-abc7-53e5c7461778
Subscriber 2 received token f7facb6f-2c16-4b0f-abc7-53e5c7461778
Subscriber 3 received token f7facb6f-2c16-4b0f-abc7-53e5c7461778

Subscriber 1 received token 67419f6a-65ff-482d-9ede-c38b1405e31e
Subscriber 2 received token 67419f6a-65ff-482d-9ede-c38b1405e31e
Subscriber 3 received token 67419f6a-65ff-482d-9ede-c38b1405e31e

Subscriber 1 received token b978b439-3ae5-440e-9d96-9f320db6e051
Subscriber 2 received token b978b439-3ae5-440e-9d96-9f320db6e051
Subscriber 3 received token b978b439-3ae5-440e-9d96-9f320db6e051

此处https://stackblitz.com/edit/rxjs-kqybmd

演示

Answer 2

好吧，只需保留一个变量即可：const val = fetchData().pipe(shareReplay(1)); 然后： val.pipe(switchMap((value) => isExpired(value) ? val : of(value)))。

Rxjs shareReplay无效（超时）

2 个答案: