AutoMapper - 继承映射不​​起作用,相同的源,多个目标

时间:2012-12-21 16:51:01

标签: c# inheritance automapper automapper-2

我是否可以在AutoMapper(v2.2)中为具有相同Source类型但目标类型不同的地图使用继承映射?

我有这种基本情况(真正的类有更多属性):

public abstract class BaseViewModel
{
    public int CommonProperty { get; set;}
}

public class ViewModelA : BaseViewModel
{
    public int PropertyA { get; set; }
}

public class ViewModelB : BaseViewModel
{
    public int PropertyB { get; set; }
}

ViewModelAViewModelB是同一个实体类的不同表示形式:

public class Entity
{
    public int Property1 { get; set; }
    public int Property2 { get; set; }
    public int Property3 { get; set; }
}

我想为每个ViewModel重用BaseViewModel的相同映射,例如:

Mapper.CreateMap<Entity, BaseViewModel>()
    .Include<Entity, ViewModelA>()
    .Include<Entity, ViewModelB>()
    .ForMember(x => x.CommonProperty, y => y.MapFrom(z => z.Property1));

Mapper.CreateMap<Entity, ViewModelA>()
    .ForMember(x => x.PropertyA, y => y.MapFrom(z => z.Property2));

Mapper.CreateMap<Entity, ViewModelB>()
    .ForMember(x => x.PropertyB, y => y.MapFrom(z => z.Property3));

但不幸的是,这似乎不起作用。这些电话是这样的:

var model = Mapper.Map<Entity, ViewModelA>(entity);

导致model映射PropertyA,但不CommonProperty。我相信我正在正确地遵循https://github.com/AutoMapper/AutoMapper/wiki/Mapping-inheritance中的示例,但我担心使用相同的Source类型创建多个地图会使AutoMapper绊倒。

任何见解?我喜欢将Base类映射分组在一起的想法,但这似乎不起作用。

2 个答案:

答案 0 :(得分:15)

不幸的是,在这种情况下,AutoMapper似乎只为每个源类型注册一个子类映射,最后一个(ViewModelB)。这可能是为了使用并行层次结构而不是单一源类型。

要解决此问题,您可以在扩展方法中封装公共映射:

public static IMappingExpression<Entity, TDestination> MapBaseViewModel<TDestination>(this IMappingExpression<Entity, TDestination> map)
  where TDestination : BaseViewModel { 
  return map.ForMember(x => x.CommonProperty, y => y.MapFrom(z => z.Property1));
}

并在单个子类映射中使用它:

Mapper.CreateMap<Entity, ViewModelA>()
  .MapBaseViewModel<ViewModelA>()
  .ForMember(x => x.PropertyA, y => y.MapFrom(z => z.Property2));

Mapper.CreateMap<Entity, ViewModelB>()
  .MapBaseViewModel<ViewModelB>()
  .ForMember(x => x.PropertyB, y => y.MapFrom(z => z.Property3));

答案 1 :(得分:0)

哟可以这样做

            CreateMap<Entity, ViewModelA>()
            .InheritMapping(x =>
            {
                x.IncludeDestinationBase<BaseViewModel>();
            });

有扩展代码

public static class MapExtensions
{        

    public static void InheritMapping<TSource, TDestination>(
        this IMappingExpression<TSource, TDestination> mappingExpression,
        Action<InheritMappingExpresssion<TSource, TDestination>> action)
    {
        InheritMappingExpresssion<TSource, TDestination> x =
            new InheritMappingExpresssion<TSource, TDestination>(mappingExpression);
        action(x);
        x.ConditionsForAll();
    }

    private static bool NotAlreadyMapped(Type sourceType, Type desitnationType, ResolutionContext r, Type typeSourceCurrent, Type typeDestCurrent)
    {
        var result = !r.IsSourceValueNull &&
               Mapper.FindTypeMapFor(sourceType, desitnationType).GetPropertyMaps().Where(
                   m => m.DestinationProperty.Name.Equals(r.MemberName)).Select(y => !y.IsMapped()
                   ).All(b => b);
        return result;
    }
    public class InheritMappingExpresssion<TSource, TDestination>
    {
        private readonly IMappingExpression<TSource, TDestination> _sourcExpression;
        public InheritMappingExpresssion(IMappingExpression<TSource, TDestination> sourcExpression)
        {
            _sourcExpression = sourcExpression;
        }
        public void IncludeSourceBase<TSourceBase>(
            bool ovverideExist = false)
        {
            Type sourceType = typeof (TSourceBase);
            Type destinationType = typeof (TDestination);
            if (!sourceType.IsAssignableFrom(typeof (TSource))) throw new NotSupportedException();
            Result(sourceType, destinationType);
        }
        public void IncludeDestinationBase<TDestinationBase>()
        {
            Type sourceType = typeof (TSource);
            Type destinationType = typeof (TDestinationBase);
            if (!destinationType.IsAssignableFrom(typeof (TDestination))) throw new NotSupportedException();
            Result(sourceType, destinationType);
        }
        public void IncludeBothBases<TSourceBase, TDestinatioBase>()
        {
            Type sourceType = typeof (TSourceBase);
            Type destinationType = typeof (TDestinatioBase);
            if (!sourceType.IsAssignableFrom(typeof (TSource))) throw new NotSupportedException();
            if (!destinationType.IsAssignableFrom(typeof (TDestination))) throw new NotSupportedException();
            Result(sourceType, destinationType);
        }
        internal void ConditionsForAll()
        {
            _sourcExpression.ForAllMembers(x => x.Condition(r => _conditions.All(c => c(r))));//указываем что все кондишены истинны
        }
        private List<Func<ResolutionContext, bool>> _conditions = new List<Func<ResolutionContext, bool>>();
        private void Result(Type typeSource, Type typeDest)
        {
                _sourcExpression.BeforeMap((x, y) =>
                {
                    Mapper.Map(x, y, typeSource, typeDest);
                });
                _conditions.Add((r) => NotAlreadyMapped(typeSource, typeDest, r, typeof (TSource), typeof (TDestination)));
        }
    }

}
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