Question

我正在尝试为板球比赛模拟生成以下值：

Overs-> 0-19（共20个overs）
球-> 1-6（共120个，每个上方有6个球）

因此，必须生成的序列类似于0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 1.1, 1.2, 1.3, ....., 19.4, 19.5, 19.6

我可以通过使用2个嵌套循环进行迭代来轻松实现此目的，但是由于我将RxJS与angular一起使用，因此我想以反应方式进行此操作。我想出了以下应该可以在AFAIK上运行的代码：

import { range } from 'rxjs';
import { map, combineAll } from 'rxjs/operators';

let overs = range(0, 20);

let balls = range(1, 6);

let ob = overs.pipe(
  map((over) => balls.pipe(
    map((ball) => `${over}.${ball}`)
  ))
).pipe(combineAll())

ob.subscribe((val) => console.log(val));

但是问题是，它缺少很多值。我得到以下输出：

[ '0.6',
  '1.6',
  '2.6',
  '3.6',
  '4.6',
  '5.6',
  '6.6',
  '7.6',
  '8.6',
  '9.6',
  '10.6',
  '11.6',
  '12.6',
  '13.6',
  '14.6',
  '15.6',
  '16.6',
  '17.6',
  '18.6',
  '19.1' ]
[ '0.6',
  '1.6',
  '2.6',
  '3.6',
  '4.6',
  '5.6',
  '6.6',
  '7.6',
  '8.6',
  '9.6',
  '10.6',
  '11.6',
  '12.6',
  '13.6',
  '14.6',
  '15.6',
  '16.6',
  '17.6',
  '18.6',
  '19.2' ]
[ '0.6',
  '1.6',
  '2.6',
  '3.6',
  '4.6',
  '5.6',
  '6.6',
  '7.6',
  '8.6',
  '9.6',
  '10.6',
  '11.6',
  '12.6',
  '13.6',
  '14.6',
  '15.6',
  '16.6',
  '17.6',
  '18.6',
  '19.3' ]
[ '0.6',
  '1.6',
  '2.6',
  '3.6',
  '4.6',
  '5.6',
  '6.6',
  '7.6',
  '8.6',
  '9.6',
  '10.6',
  '11.6',
  '12.6',
  '13.6',
  '14.6',
  '15.6',
  '16.6',
  '17.6',
  '18.6',
  '19.4' ]
[ '0.6',
  '1.6',
  '2.6',
  '3.6',
  '4.6',
  '5.6',
  '6.6',
  '7.6',
  '8.6',
  '9.6',
  '10.6',
  '11.6',
  '12.6',
  '13.6',
  '14.6',
  '15.6',
  '16.6',
  '17.6',
  '18.6',
  '19.5' ]
[ '0.6',
  '1.6',
  '2.6',
  '3.6',
  '4.6',
  '5.6',
  '6.6',
  '7.6',
  '8.6',
  '9.6',
  '10.6',
  '11.6',
  '12.6',
  '13.6',
  '14.6',
  '15.6',
  '16.6',
  '17.6',
  '18.6',
  '19.6' ]

正如我们可以看到的那样，所有的溢出都已生成（0-19），但是每个都只得到.6值。只有倒数第二个，即19th具有唯一值19.1, 19.2...

此代码中我缺少什么？

此外，如果有帮助，如果您想在浏览器中尝试使用上述代码，请参见以下代码：https://repl.it/@ishankhare07/cricket-overs

Answer 1

您甚至不需要使用combineAll而是使用mergeMap来简化理解。

import { range } from 'rxjs';
import { map, mergeMap } from 'rxjs/operators';

let overs = range(0, 20);
let balls = range(1, 6);

let ob = overs.pipe(
  mergeMap((over) => balls.pipe(
    map((ball) => `${over}.${ball}`),
  )),
); 

ob.subscribe((val) => console.log(val));

实时演示：https://stackblitz.com/edit/rxjs-aud3ot

结合使用2个Observables时RxJS缺少值

1 个答案: