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我遇到了错误处理程序的问题。我正在尝试在我的服务中创建一个通用的 retryPipeline：当调用失败时，它会在抛出和错误之前重试 3 次。到现在为止还挺好。如果我将代码放在方法中，它会起作用，如下所示：

  getMun(id_del: number, id_muno: number): Observable<Mun> {

    let urlAPIMun = urlAPI;
    urlAPIMun += '/' + id_del + '/mun' + '/' + id_mun + '?flag_geometry=true';
    return this._http.get<Mun>(urlAPIMunicipios).pipe(
     //   tap(() => console.log('HTTP request executed')),
      retryWhen(errors => errors.pipe(
        // Concat map to keep the errors in order and make sure they
        // aren't executed in parallel
        concatMap((e: HttpErrorResponse, i) =>
          // Executes a conditional Observable depending on the result
          // of the first argument
          iif(
            () => i >= 3,
            // If the condition is true we throw the error (the last error)
            throwError(e.error.errores),
            // Otherwise we pipe this back into our stream and delay the retry
            of(e).pipe(delay(5000))
          )
        ))));
  }

我试图提取管道内的代码来声明一个常量，然后在我的服务调用中调用常量：

const RETRYPIPELINE =
  retryWhen((errors) =>
    errors.pipe(
      concatMap((e: HttpErrorResponse, i) =>
          () => i >= 3,
          throwError(e.error.errores),
          of(e).pipe(delay(5000))
        )
      )
    )
  );

return this._http.get<Mun>(urlAPIMunicipios).pipe(RETRYPIPELINE);

但我收到此错误：

错误 TS2322：类型 'Observable<{}>' 不可分配给类型 'Observable'。类型“{}”缺少类型“Mun”的以下属性：id_mun、id_del、den

有什么方法可以创建一个可以分配给任何方法的通用 const，尽管该方法返回一个类型化的值？提前致谢

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1 回答 1

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最后，感谢这个答案，我修复了它：

添加演员表retryWhen，彻底解决我的问题：

export const RETRYPIPELINE =
  retryWhen<any>((errors) =>
    errors.pipe(
      // Use concat map to keep the errors in order and make sure they
      // aren't executed in parallel
      concatMap((e: HttpErrorResponse, i) =>
        // Executes a conditional Observable depending on the result
        // of the first argument
        iif(
          () => i >= 3,
          // If the condition is true we throw the error (the last error)
          throwError(e.error.errores),
          // Otherwise we pipe this back into our stream and delay the retry
          of(e).pipe(delay(5000))
        )
      )
    )
  )

;

于 2019-04-10T11:01:38.707 回答