存根或模拟IMapper返回派生类,其中基地预期
我有一个类,它在一个IMapper
在构造这样存根或模拟IMapper返回派生类,其中基地预期
public Foo(IMapper mapper)
在代码Foo的我有此线
var dao = _mapper.Map<BaseDAO>(obj);
BaseDAO
有3个子类型,在我已经设置的实际代码中这样
CreateMap<Base, BaseDAO>() .Include<Child1, Child1DAO>()
.Include<Child2, Child2DAO>()
.Include<Child3, Child3DAO>();
我想模拟出上述行
var dao = _mapper.Map<BaseDAO>(obj);
因此如果Child1
在随后被传递一个Child1DAO
将返回与同为其他亚型。我试图存根出IMapper
但下面的方法返回一个错误,指出
Child1DAO不能被隐式转换为TDestination
,我试图模拟出IMapper
但未能得到这两种工作。
public TDestination Map<TDestination>(object source) {
return new Child1DAO();
}
任何想法?
回答:
对于这个例子的目的,假设以下类是下测试受试者
public class Foo { private IMapper mapper;
public Foo(IMapper mapper) {
this.mapper = mapper;
}
public BaseDAO Bar(object obj) {
var dao = mapper.Map<BaseDAO>(obj);
return dao;
}
}
凡IMapper
依赖性具有以下合同定义
public interface IMapper { /// <summary>
/// Execute a mapping from the source object to a new destination object.
/// The source type is inferred from the source object.
/// </summary>
/// <typeparam name="TDestination">Destination type to create</typeparam>
/// <param name="source">Source object to map from</param>
/// <returns>Mapped destination object</returns>
TDestination Map<TDestination>(object source);
//...
}
下面的试验证明了,使用起订,
模拟IMapper返回派生类,其中基地期望d
[TestClass] public class TestClass {
[TestMethod]
public void _TestMethod() {
//Arrange
var mock = new Mock<IMapper>();
var foo = new Foo(mock.Object);
mock
//setup the mocked function
.Setup(_ => _.Map<BaseDAO>(It.IsAny<object>()))
//fake/stub what mocked function should return given provided arg
.Returns((object arg) => {
if (arg != null && arg is Child1)
return new Child1DAO();
if (arg != null && arg is Child2)
return new Child2DAO();
if (arg != null && arg is Child3)
return new Child3DAO();
return null;
});
var child1 = new Child1();
//Act
var actual = foo.Bar(child1);
//Assert
Assert.IsNotNull(actual);
Assert.IsInstanceOfType(actual, typeof(BaseDAO));
Assert.IsInstanceOfType(actual, typeof(Child1DAO));
}
}
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