Пример #1
0
        public void AddTwoListenersAndPerformLogSuccessfully()
        {
            //Arrange
            var asyncListener = new AsynchronousImpressionListener();
            var listener1     = new TestListener();
            var listenerMock2 = new Mock <IImpressionListener>();

            asyncListener.AddListener(listener1);
            asyncListener.AddListener(listenerMock2.Object);


            //Act
            asyncListener.Log(new KeyImpression()
            {
                feature = "test", changeNumber = 100, keyName = "date", label = "testdate", time = 10000000, treatment = "on", bucketingKey = "any"
            });
            Thread.Sleep(1000);

            //Assert
            Assert.AreEqual(listener1.list.Count, 1);
            listenerMock2.Verify(x => x.Log(It.Is <KeyImpression>(p => p.keyName == "date" && p.feature == "test" && p.treatment == "on" && p.time == 10000000 && p.changeNumber == 100 && p.label == "testdate" && p.bucketingKey == "any")), Times.Once());
        }
Пример #2
0
        public void AddTwoListenersAndPerformLogSuccessfullyWithOneLongRunningTask()
        {
            //Arrange
            var asyncListener = new AsynchronousImpressionListener();
            var listener1     = new TestListener2();
            var listener2     = new TestListener();

            asyncListener.AddListener(listener1);
            asyncListener.AddListener(listener2);


            //Act
            asyncListener.Log(new KeyImpression()
            {
                feature = "test", changeNumber = 100, keyName = "date", label = "testdate", time = 10000000, treatment = "on", bucketingKey = "any"
            });
            Thread.Sleep(1000);

            //Assert
            Assert.AreEqual(listener2.list.Count, 1);
            Thread.Sleep(10000);
            Assert.AreEqual(listener1.list.Count, 1);
        }