Angualr / RXJS跟踪处于重试模式的HTTP请求

问题描述 投票:1回答:1

我有一个为我处理多个API请求的函数,如果失败,则将每个请求置于重试模式。现在,如果一个请求已经在重试循环中,并且同一API调用的另一个实例到来,我的函数将无法跟踪此请求,并再次在重试循环中添加冗余的API调用。

Assuming i am placing a call to
/api/info/authors

What is happening

1stREQ| [re0]------>[re1]------>[re2]------>[re3]------>[re4]------>[re5]
2ndREQ|                         [re0]------>[re1]------>[re2]------>[re3]------>[re4]------>[re5]


What should happen,

1stREQ| [re0]------>[re1]------>[re2]------>[re3]------>[re4]------>[re5]
2ndREQ|                         [re0]/ (MERGE)

以下是我的服务有我的重试功能,

import { Injectable } from '@angular/core';
import { HttpClient, HttpParams } from '@angular/common/http';
import { retryWhen, mergeMap, finalize, share, shareReplay } from 'rxjs/operators';
import { Observable, throwError, of, timer } from 'rxjs';


@Injectable({
  providedIn: 'root'
})
export class DataService {
  constructor(private http: HttpClient) { }

  private apiUrl:string = 'http://localhost/api-slim-php/public/api';
  public dataServerURL:string = 'http://localhost/';

/* 
This function fetches all the info from API /info/{category}/{id}
category  : author    & id  : '' or 1,2,3... or a,b,c...
category  : form      & id  : '' or 1,2,3...
category  : location  & id  : '' or 1,2,3...
category  : school    & id  : '' or 1,2,3...
category  : timeframe & id  : '' or 1,2,3...
category  : type      & id  : '' or 1,2,3...
 */            
  public getInfoAPI(category:string, id:string = "", page:string = "1", limit:string = "10"){
    var callURL : string = '';

    if(!!id.trim() && !isNaN(+id)) callURL = this.apiUrl+'/info/'+category+'/'+id;
    else callURL = this.apiUrl+'/info/'+category;

    return this.http.get(callURL,{
      params: new HttpParams()
        .set('page', page)
        .set('limit', limit)
    }).pipe(
      retryWhen(genericRetryStrategy({maxRetryAttempts: 5, scalingDuration: 1000})),
      shareReplay()
    );
  }
 }
export const genericRetryStrategy = ({
  maxRetryAttempts = 3,
  scalingDuration = 1000,
  excludedStatusCodes = []
}: {
  maxRetryAttempts?: number,
  scalingDuration?: number,
  excludedStatusCodes?: number[]
} = {}) => (attempts: Observable<any>) => {
  return attempts.pipe(
    mergeMap((error, i) => {
      const retryAttempt = i + 1;
      // if maximum number of retries have been met
      // or response is a status code we don't wish to retry, throw error
      if (
        retryAttempt > maxRetryAttempts ||
        excludedStatusCodes.find(e => e === error.status)
      ) {
        console.log(error);
        return throwError(error);
      }
      console.log(
        `Attempt ${retryAttempt}: retrying in ${retryAttempt *
          scalingDuration}ms`
      );
      // retry after 1s, 2s, etc...
      return timer(retryAttempt * scalingDuration);
    }),
    finalize(() => console.log('We are done!'))
  );
};

注意:

有些人可能建议我执行shareReplay(),但它无法处理来自其他两个组件/源的相同请求

以下应仅为6,相反,在快速单击调用相同API的两个按钮时,它们为12。 (缩放持续时间为1000ms)

enter image description here

注:

请避免使用FLAGS,在我看来这是最终的威胁。

angular rxjs observable httprequest
1个回答
0
投票

注意,每次调用getInfoAPI() http.get()创建一个新的可观察对象,并且shareReplay()共享该新的可观察对象,它不会合并两个调用。如果希望呼叫者获得合并的可观测值,则可以从两个调用中返回相同的可观测值。但这是错误的解决方案,我将在后面解释。例如:

export class DataService {
  private readonly getInfoRequest = new Subject<GetInfoRequest>();
  private readonly getInfoResponse = this.getInfoRequest.pipe(
    exhaustMap(request => { 
      const callURL = createGetInfoUrl(request);
      const callParams = createGetInfoParams(request);
      return this.http.get(callURL, callParams).pipe(
        retryWhen( ... );
      );
    })
  );

  public getInfoAPI(category:string, id:string = "", page:string = "1", limit:string = "10") {
    this.getInfoRequest.next({ category: category, id: id, page: page, limit: limit });
    return this.getInfoResponse;
  }

  ...
}

上面的代码具有与您尝试通过shareReplay()实现的功能相同的功能,但是如果调用参数不匹配怎么办?一个组件已请求第一页,但另一个组件已请求第二页,第二个组件将接收第一页而不是第二个页面。因此,我们也应该考虑调用参数,事情变得更加复杂。

一种解决方案是用存储库包装HttpService,该存储库将处理缓存,它可以将数据缓存在内存中,数据库内部或其他地方,但是我怀疑这就是您想要的。据我了解,问题是并发请求,更好的方法是阻止此类请求。例如,如果某个请求是通过单击按钮触发的,则只需在执行请求时禁用该按钮,或跳过重复的请求即可。这是解决此类问题的常用方法。

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