如何用静态方法来调用静态类（UnitOfWork case）？

我有这些类：如何用静态方法来调用静态类（UnitOfWork case）？

public static class UnitOfWorkSS 
{ 
    public static IUnitOfWork Begin() 
    { 
    return IoC.Resolve<IUnitOfWork>(); 
    } 
} 

public class PostService 
{ 
    using (IUnitOfWork unitOfWork = UnitOfWorkSS.Begin()) 
    { 
    //don't forget to sanitize html content 
    htmlContent = _htmlSanitizer.Sanitize(htmlContent); 

    IPost post = _factory.CreatePost(byUser, title, htmlContent);      

    _postRepository.Add(post); 

    unitOfWork.Commit(); 
    } 
}

我怎么能嘲笑类UnitOfWorkSS和unitOfWork？

来源

2010-11-26 Attilah

它看起来像你与呼叫做开始（）将返回您配置类特定接口的唯一的事情：IUnitOfWork

你真的只需要确保你的电话开始（）返回模拟实现的IUnitOfWork

一两种方法可以做到这一点：

方案一 - 重构UnitOfWorkSS，使您可以设置实例将返回

public static class UnitOfWorkSS 
{ 
    private static IUnitOfWork _unitOfWork; 
    public static IUnitOfWork UnitOfWork 
    { 
     set { _unitOfWork = value; } 
     private get{ _unitOfWork ?? (_unitOfWork = IoC.Resolve<IUnitOfWork>()); } 
    } 

    public static IUnitOfWork Begin() 
    { 
     return UnitOfWork; 
    } 
} 

[TestMethod] 
public void DoStuff() 
{ 
    var mockUnitOfWork = new Mock<IUnitOfWork>(); 
    UnitOfWorkSS.UnitOfWork = mockUnitOfWork.Object; 

    //Do some setup and verify 
}

方案二 - 只要你的IoC容器注册的IUnitOfWork一个模拟实例

private Mock<IUnitOfWork> _mockUnitOfWork; 

[TestInitialize] 
public void Init() 
{ 
    _mockUnitOfWork = new Mock<IUnitOfWork>(); 

    //Making a lot of assumptions about your IoC here... 
    IoC.Register<IUnitOfWork>(_mockUnitOfWork.Object); 
} 

[TestMethod] 
public void DoStuff() 
{ 
    _mockUnitOfWork.Setup(...); 

    //Do some verification 
}

来源

2010-11-26 04:23:40 Josh