Question

我想要模拟懒惰的界面，但我得到object reference not set to an instance of an object例外。

这是受测试的课程：

public class ProductServiceService : IProductServiceService
{
    private readonly Lazy<IProductServiceRepository> _repository;
    private readonly Lazy<IProductPackageRepository> _productPackageRepository;

    public ProductServiceService(
        Lazy<IProductServiceRepository> repository,
        Lazy<IProductPackageRepository> productPackageRepository)
    {
        _repository = repository;
        _productPackageRepository = productPackageRepository;
    }

    public async Task<OperationResult> ValidateServiceAsync(ProductServiceEntity service)
    {
        var errors = new List<ValidationResult>();

        if (!await _productPackageRepository.Value.AnyAsync(p => p.Id == service.PackageId))
            errors.Add(new ValidationResult(string.Format(NameMessageResource.NotFoundError, NameMessageResource.ProductPackage)));

       .
       .
       .

        return errors.Any()
            ? OperationResult.Failed(errors.ToArray())
            : OperationResult.Success();
    }
}

这是测试类

[Fact, Trait("Category", "Product")]
public async Task Create_Service_With_Null_Financial_ContactPerson_Should_Fail()
{
    // Arrange
    var entity = ObjectFactory.Service.CreateService(packageId: 1);

    var fakeProductServiceRepository = new Mock<Lazy<IProductServiceRepository>>();

    var repo= new Mock<IProductPackageRepository>();
    repo.Setup(repository => repository.AnyAsync(It.IsAny<Expression<Func<ProductPackageEntity, bool>>>()));
    var fakeProductPackageRepository  = new Lazy<IProductPackageRepository>(() => repo.Object);

    var sut = new ProductServiceService(fakeProductServiceRepository.Object, fakeProductPackageRepository);

    // Act
    var result = await sut.AddServiceAsync(service);

    // Assert
    Assert.False(result.Succeeded);
    Assert.Contains(result.ErrorMessages, error => error.Contains(string.Format(NameMessageResource.NotFoundError, NameMessageResource.ProductPackage)));
}

fakeProductPackageRepository始终为空。我关注了这篇博文，但我仍然得到了空引用异常。

How to mock lazy initialization of objects in C# unit tests using Moq

更新：这是一个表示fakeProductPackageRepository为空的屏幕。

Answer 1

以下是您的示例的重构版本：

[Fact, Trait("Category", "Product")]
public async Task Create_Service_With_Null_Financial_ContactPerson_Should_Fail() {
    // Arrange
    var entity = ObjectFactory.Service.CreateService(packageId = 1);

    var productServiceRepositoryMock = new Mock<IProductServiceRepository>();

    var productPackageRepositoryMock = new Mock<IProductPackageRepository>();
    productPackageRepositoryMock
        .Setup(repository => repository.AnyAsync(It.IsAny<Expression<Func<ProductPackageEntity, bool>>>()))
        .ReturnsAsync(false);

    //Make use of the Lazy<T>(Func<T>()) constructor to return the mock instances
    var lazyProductPackageRepository = new Lazy<IProductPackageRepository>(() => productPackageRepositoryMock.Object);
    var lazyProductServiceRepository = new Lazy<IProductServiceRepository>(() => productServiceRepositoryMock.Object);

    var sut = new ProductServiceService(lazyProductServiceRepository, lazyProductPackageRepository);

    // Act
    var result = await sut.AddServiceAsync(service);

    // Assert
    Assert.False(result.Succeeded);
    Assert.Contains(result.ErrorMessages, error => error.Contains(string.Format(NameMessageResource.NotFoundError, NameMessageResource.ProductPackage)));
}

<强>更新

经过测试后，您所述问题的Minimal, Complete, and Verifiable example 通过。

[TestClass]
public class MockLazyOfTWithMoqTest {
    [TestMethod]
    public async Task Method_Under_Test_Should_Return_True() {
        // Arrange
        var productServiceRepositoryMock = new Mock<IProductServiceRepository>();

        var productPackageRepositoryMock = new Mock<IProductPackageRepository>();
        productPackageRepositoryMock
            .Setup(repository => repository.AnyAsync())
            .ReturnsAsync(false);

        //Make use of the Lazy<T>(Func<T>()) constructor to return the mock instances
        var lazyProductPackageRepository = new Lazy<IProductPackageRepository>(() => productPackageRepositoryMock.Object);
        var lazyProductServiceRepository = new Lazy<IProductServiceRepository>(() => productServiceRepositoryMock.Object);

        var sut = new ProductServiceService(lazyProductServiceRepository, lazyProductPackageRepository);

        // Act
        var result = await sut.MethodUnderTest();

        // Assert
        Assert.IsTrue(result);
    }

    public interface IProductServiceService { }
    public interface IProductServiceRepository { }
    public interface IProductPackageRepository { Task<bool> AnyAsync();}

    public class ProductServiceService : IProductServiceService {
        private readonly Lazy<IProductServiceRepository> _repository;
        private readonly Lazy<IProductPackageRepository> _productPackageRepository;

        public ProductServiceService(
            Lazy<IProductServiceRepository> repository,
            Lazy<IProductPackageRepository> productPackageRepository) {
            _repository = repository;
            _productPackageRepository = productPackageRepository;
        }

        public async Task<bool> MethodUnderTest() {
            var errors = new List<ValidationResult>();

            if (!await _productPackageRepository.Value.AnyAsync())
                errors.Add(new ValidationResult("error"));

            return errors.Any();
        }
    }
}

Answer 2

作为参数的Lazy<>有些出乎意料，虽然不是非法的（显然）。请记住，围绕服务的Lazy<>实际上只是延迟执行Factory方法。为什么不将工厂传递给构造函数？您仍然可以在实施类中的Lazy<>中将调用包装到工厂，但是您可以在测试中假冒/嘲笑工厂并将其传递给sut。

或者，或许你在Lazy<>附近传递的原因是因为你真的在处理一个单身人士。在这种情况下，我仍然会创建一个工厂并依赖于IFactory<>。然后，工厂实现可以在其中包含Lazy<>。

通常，我通过在IoC容器中为依赖项设置自定义对象作用域来解决单例要求（没有延迟加载）。例如，StructureMap可以轻松地将某些依赖项设置为Web应用程序中的单例或每个请求范围。

我很少需要断言我在被测系统内的某些服务上做了一个懒惰的初始化。我可能需要验证我是否仅在某个范围内初始化了一次服务，但仍然可以通过伪造工厂界面来轻松测试。

Answer 3

问题是你正在创建一个懒惰的模拟作为fakeProductServiceRepository，稍后将返回那个只需要模拟的实例。

你应该改变

var fakeProductServiceRepository = new Mock<Lazy<IProductServiceRepository>>();

到

var fakeProductServiceRepository = new Mock<IProductServiceRepository>();

与Moq模拟懒惰的界面

3 个答案: