Question

我正在编写单元测试，其中SUT（被测系统）可以进行各种不同的配置。我决定（或尝试）采取另一种更清洁的方法，而不是为每种组合添加参数。对于那些不了解的人来说，Fixture对象来自一个名为AutoFixture的库，它可以处理播种随机测试值。

这是我失败的测试 -

    [Test]
    public void Test()
    {
        var locate = Build(x => x.With(xx => xx.TicketNo, 123));

        Assert.AreEqual(123, locate.TicketNo);
    }

    private Locate Build(Action<ICustomizationComposer<Locate>> customizationAction)
    {
        var fixture = new Fixture();
        var customizationComposer = fixture.Build<Locate>();
        customizationAction(customizationComposer);
        var postProcessComposer = customizationComposer
            .Without(x => x.Attachments)
            .Without(x => x.Comments)
            .Without(x => x.Reviews)
            .Without(x => x.ScheduledCrew)
            .Without(x => x.PendingDecision)
            .Without(x => x.FinalDecision)
            .Without(x => x.ConflictResolution);

        return postProcessComposer.Create();
    }

尽可能地，我尝试使用Build方法通过Locate方法调用创建一组带有一组忽略规则的Without对象。这是我每次测试所需的基线对象。每个测试都可以通过With方法调用忽略其他字段或硬代码字段值。

我希望通过，因为我设置了Build lambda参数来将TicketNo属性设置为123.相反，测试失败，因为AutoFixture将TicketNo种子随机整数。< / p>

我怎样才能做到这一点？

Answer 1

的解决方案！

    private Locate Build(Func<IPostprocessComposer<Locate>, IPostprocessComposer<Locate>> action)
    {
        var customizationComposer = _fixture.Build<Locate>();
        var postProcessComposer = customizationComposer
            .With(x => x.TicketNo, DateTime.Now.ToFileTime())
            .With(x => x.On1CallNotified, false)
            .Without(x => x.Attachments)
            .Without(x => x.Comments)
            .Without(x => x.Reviews)
            .Without(x => x.ScheduledCrew)
            .Without(x => x.PendingDecision)
            .Without(x => x.FinalDecision)
            .Without(x => x.ConflictResolution);

        postProcessComposer = action(postProcessComposer);

        return postProcessComposer.Create();
    }

如何通过参数执行其他对象配置？

1 个答案: