RxJS

Question

我想为此逐渐增加延迟：

const source = from(839283, 1123123, 63527, 4412454); // note: this is random
const spread = source.pipe(concatMap(value => of(value).pipe(delay(1000)))); // how to incrementally increase the delay where the first item emits in a second, the second item emits in three seconds, the third item emits in five seconds, and the last item emits in seven seconds.
spread.subscribe(value => console.log(value));

我知道使用间隔来逐步增加延迟时间，如下所示。但是我还需要消耗这个来源const source = from(839283, 1123123, 63527, 4412454);

const source = interval(1000); // But I also need to use have this from(839283, 1123123, 63527, 4412454)
const spread = source.pipe(concatMap(value => of(value).pipe(delay(value * 200))));
spread.subscribe(value => console.log(value

如何在 const source = from(839283, 1123123, 63527, 4412454); 开始时逐渐增加延迟时间？

Answer 1

您可以使用发射值的索引并据此计算延迟。

concatMap((value, index) => {
  return of(value).pipe(delay((index > 2 ? 7 : index) * 1000));
})

这是一个包含您的代码段的完整示例的 stackblitz： https://stackblitz.com/edit/rxjs-jjmlta?devtoolsheight=60

RxJS - 如何在不使用间隔的情况下逐步增加延迟时间？

RxJS - How to incrementally increase the delay time without using interval?

rxjs5

angular

rxjs6

rxjs-observables