嘲笑阿卡的儿童演员

Question

我正在尝试为我的演员编写单元测试，并且我坚持基本的嘲弄。 PriceAggregateActor正在使用akka持久性，我不想传递所有的conf，并想完全嘲笑它。

这是我要测试的演员

object CommandPriceActor {
  def apply() = Props(classOf[CommandPriceActor], PriceAggregateActor())
}

class CommandPriceActor(priceAggregateActorProps: Props) extends Actor with ActorLogging {

  val priceAggregateActor = context.actorOf(priceAggregateActorProps, "priceAggregateActor")

因此，在我的测试中，我尝试做类似的事情：

class CommandPriceActorTest extends TestKit(ActorSystem("test-benefits",
  ConfigFactory.parseString("""akka.loggers = ["akka.testkit.TestEventListener"] """))) with FlatSpecLike with Matchers
  with BeforeAndAfterAll with Eventually{

  class MockedChild extends Actor {
    def receive = {
      case _ => lala
    }
  }

  val probe = TestProbe()
  val commandPriceActor = TestActorRef(new CommandPriceActor(Props[MockedChild]))

我总是得到：

Caused by: java.lang.IllegalArgumentException: no matching constructor found on class CommandPriceActorTest$MockedChild for arguments []

为什么抱怨mockedChild？它不应该采用任何构造函数参数。

Answer 1

这是因为MockedChild是您测试的儿童演员。缺少的构造函数参数是对测试的引用（它是父类）。

您有三种选择：

1. 将对this的引用传递给Props
2. 使用Props
的命名参数形式
3. 使MockedChild成为顶级类（或对象的成员）

选项1

val probe = TestProbe()
val mockProps = Props(classOf[MockedChild], this)
val commandPriceActor = TestActorRef(new CommandPriceActor(mockProps))

选项2

val probe = TestProbe()
val mockProps = Props(new MockedChild)
val commandPriceActor = TestActorRef(new CommandPriceActor(mockProps))

选项3

val probe = TestProbe()
val mockProps = Props(new CommandPriceActorTest.MockedChild)
val commandPriceActor = TestActorRef(new CommandPriceActor(mockProps))

// ....

object CommandPriceActorTest {
  class MockedChild extends Actor {
    def receive = {
      case _ => lala
    }
  }
}