ScalaTest AsyncFunSuiteLike多个断言

Fed*_* E. 3 scala scalatest

我正在重写用于测试以String响应的Akka演员的测试,现在,我们使用的函数返回a Future[String]。因此,请使用AsyncFunSuiteLike重写测试。

目前,我正在测试一个必须断言多个Future的测试,但是我似乎不太了解如何使用AsyncFunSuiteLike实现它。

我的测试如下所示:

test("Error responses") {

    NanoHTTPD.Response.Status.values().filter(status => status.getRequestStatus >= 400).map {
      status => {
        val statusCode = status.getRequestStatus
        httpService.setStatusCode(status)
        val responseBody = s"Request failed with status $status"
        httpService.setResponseContent(responseBody)
        val errorMessage = s"Error response (${status.getRequestStatus}): $responseBody"

        myobject.request("123456").map {
          resp => assert(resp === "....")
        }
      }
    }

}
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但是我遇到一个错误 type missmatch. Expected: Future[Assertion] actual: List[Future[scalatest.Assertion]]

关于如何在单个测试中实现多个断言的想法?

Mar*_*lic 6

ScalaTest 异步测试期望测试主体返回Future[Assertion]。现在您的测试体返回,List[Future[Assertion]]]因为

NanoHTTPD.Response.Status.values() ... // Returns List
  ...
    myobject.request("123456").map(resp => assert(resp === "....")} // Returns Future[Assertion]
  ...
}
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我们可以转换List[Future[T]]Future[List[T]]使用

Future.sequence(listOfFutures)
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并且,我们可以转换List[Assertion]Assertion使用

assert(listOfAssertions.forall(_ == Succeeded))
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放在一起我们得到

Future.sequence {
  NanoHTTPD.Response.Status.values().filter(status => status.getRequestStatus >= 400).map {
    status => {
      val statusCode = status.getRequestStatus
      httpService.setStatusCode(status)
      val responseBody = s"Request failed with status $status"
      httpService.setResponseContent(responseBody)
      val errorMessage = s"Error response (${status.getRequestStatus}): $responseBody"

      myobject.request("123456").map {
        resp => assert(resp === "....")
      }
    }
  }
}.map(listOfAssertions => assert(listOfAssertions.forall(_ == Succeeded)))
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