Question

我在错误处理程序方面遇到问题。我正在尝试在服务中创建一个通用的retryPipeline：当调用失败时，它将在抛出错误之前重试3次。到目前为止，一切都很好。如果将代码放入方法中，则可以正常工作，

  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))
          )
        ))));
  }

我试图在管道中引出代码以声明一个const，然后在我的服务调用中调用const：

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);

但是我收到此错误：

错误TS2322：类型'Observable <{}>'无法分配给类型'Observable'。类型“ {}”缺少类型“ Mun”中的以下属性：id_mun，id_del，den

有什么方法可以创建一个通用的const，尽管该方法返回一个类型化的值，但它可以赋值给任何方法？预先感谢

Answer 1

最后，我感谢this answer：

在retryWhen中添加演员表，可以彻底解决我的问题：

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))
        )
      )
    )
  )

;

在const angular上创建通用类型

1 个答案: