RxJS-如何在不使用间隔的情况下逐步增加延迟时间？

问题描述

我想为此增加延迟：

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渐增加延迟时间，如下所示。但我还需要使用此来源const source = from(839283,4412454);

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

从const source = from(839283,4412454); 开始时如何逐步增加延迟时间？

解决方法

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));
})

以下是有关代码段的完整示例的概述： https://stackblitz.com/edit/rxjs-jjmlta?devtoolsheight=60