RXJS:自适应超时

Question

我正在使用 RXJS 6，我尝试使用自适应超时进行重试(我第一次尝试 1s 超时，如果不成功，我会重试但 2s 超时，等等......

看起来重试时超时值没有更新，我是 RxJS 的新手，我找不到问题。

这里是代码片段：

import { zip, timer, range, of, pipe } from 'rxjs';
import {  tap, retry, concatMap, timeout, catchError, delay, retryWhen } from 'rxjs/operators';
let timeoutFactor = 1;
let timeoutInit = 2000;
let timeoutValue = timeoutInit;
// simulate request
function makeRequest(timeToDelay) {  
return of('Request Complete! '+timeToDelay).pipe(
tap( () => console.log(`doing request (${timeToDelay}ms) with timeout of ${timeoutValue}`)),
delay(timeToDelay));
}

const obs$ = of(4000, 3000, 2000)
.pipe( 
concatMap(duration => makeRequest(duration).pipe(
timeout(timeoutValue),
retryWhen(error => zip(error, range(1,3).pipe(
tap( i  => {
console.log("set timeout on " + (i*1000) + " ms")
setTimeoutFactor(i);
})
)))    
))
);
obs$.subscribe(
val => console.log(val),
err => console.log(`err: ${err}` ),
() => console.log(`complete` ));
function setTimeoutFactor(value){
timeoutValue = timeoutInit*value;
}

但我的结果并不是很好：

doing request (4000ms) with timeout of 2000
set timeout on 1000 ms
set timeout on 2000 ms
set timeout on 3000 ms
doing request (4000ms) with timeout of 6000
doing request (4000ms) with timeout of 6000
doing request (4000ms) with timeout of 6000
doing request (3000ms) with timeout of 6000
timeout: 6000
Request Complete! 3000
doing request (2000ms) with timeout of 6000
timeout: 6000
Request Complete! 2000
complete

有什么帮助吗？

Answer 1

在您选择的延迟后重试不会那么复杂。例如，以下(经过测试的工作代码(示例在指数延迟(1s、4s、9s、16s、25s ...(后重试：

// source observable simply throws error, which is ideal for testing retry concept
const source = throwError(new Error('Oops..!'));
const smartSource = source.pipe(retryWhen(error => {
return error.pipe(mergeMap((e, i) => {
const delayFactor = i * i *  1000;
console.log(`Retrying after ${delayFactor / 1000} seconds`);
return of('Retry signal').pipe(delay(delayFactor));
}));
}));
smartSource.subscribe(x => console.log('result...'));

如果要更改延迟逻辑，只需根据需要修改delayFactor即可。例如，如果要每秒重试一次：

const delayFactor = i * 1000;