Question

我发现ProjectTo可以为导航属性生成不同的对象。我会举例说明。

我的模特：

    public class Element 
    {
        public Guid Id { get; set; }
        public string Name { get; set;}
    }

    public class CircuitElement
    {
        public Guid Id { get; set; }
        public Guid ElementId { get; set; }
        public Guid CircuitId { get; set; }
        public Element Element { get; set;}
    }

当我仅使用EF.Core请求时，我在每个CircuitElement的导航属性中都得到了完全相同的Element对象：

    List<CircuitElement> circuitElements = UnitOfWork.GetRepository<CircuitElement>()
        .GetQuery(csc => csc.CircuitId == circuitId)
        .Include(x => x.Element)
        .ToList();

    var elements = circuitElements.Select(x => x.Element).Where(x => x.Id == GuidConst.AxElement).ToList();
    var comp1 = ReferenceEquals(elements[0], elements[1]); //returns true
    var comp2 = ReferenceEquals(elements[0], elements[2]); //returns true

然后我要添加DTO，MappingProfile和ProjectTo：

    public class ElementDto
    {
        public Guid Id { get; set; }
        public string Name { get; set;}
    }

    public class CircuitElementDto
    {
        public Guid Id { get; set; }
        public Guid ElementId { get; set; }
        public Guid CircuitId { get; set; }
        public ElementDto Element { get; set;}
    }

    public class MappingProfile : Profile
    {
        public MappingProfile()
        {
            CreateMap<Element, ElementDto>(MemberList.None);
            CreateMap<ElementDto, Element>(MemberList.None);

            CreateMap<CircuitElement, CircuitElementDto>(MemberList.None);
            CreateMap<CircuitElementDto, CircuitElement>(MemberList.None);
        }
    }

    List<CircuitElementDto> circuitElements = UnitOfWork.GetRepository<CircuitElement>()
        .GetQuery(csc => csc.CircuitId == circuitId)
        .Include(x => x.Element)
        .ProjectTo<CircuitElementDto>(Mapper.ConfigurationProvider)
        .ToList();

    var elements2 = circuitElements.Select(x => x.Element).Where(x => x.Id == GuidConst.AxElement).ToList();
    var comp3 = ReferenceEquals(elements[0], elements[1]); //returns FALSE
    var comp4 = ReferenceEquals(elements[0], elements[2]); //returns FALSE

如您所见，这些元素是不同的对象。是否可以配置Automapper，所以它将为所有这些导航生成一个对象，而不是3个？还是在ToList之后有合并它们的技术？

Automapper Project为导航属性不生成ReferenceEquals对象

0 个答案: