我在映射包含继承自IEnumerable
的自定义列表的属性时遇到问题(如果我删除了该继承,则此示例有效)。我已将问题简化为此模型:
public interface IMyEnumerable<T> : IEnumerable<T> { }
public class MyIEnumerable<T> : IMyEnumerable<T>
{
private readonly IEnumerable<T> _items;
public MyIEnumerable(IEnumerable<T> items)
{
_items = items;
}
public IEnumerator<T> GetEnumerator()
{
return _items.GetEnumerator();
}
IEnumerator IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}
public class Source
{
public List<SourceItem> Items { get; set; }
}
public class Destination
{
public IMyEnumerable<DestinationItem> Items { get; set; }
}
public class SourceItem
{
public string Name { get; set; }
}
public class DestinationItem
{
public string Name { get; set; }
}
然后我尝试使用这种方式:
public class MyResolver : ValueResolver<Source, IMyEnumerable<DestinationItem>>
{
protected override IMyEnumerable<DestinationItem> ResolveCore(Source source)
{
var destinationItems = Mapper.Map<List<SourceItem>, IEnumerable<DestinationItem>>(source.Items);
return new MyIEnumerable<DestinationItem>(destinationItems);
}
}
// Mappings
Mapper.CreateMap<Source, Destination>()
.ForMember(x => x.Items, m => m.ResolveUsing<MyResolver>());
Mapper.CreateMap<SourceItem, DestinationItem>();
// Using the mappings
var source = // not really relevant
var destination = Mapper.Map<Destination>(source);
这给了我以下异常(为了便于阅读而略微编辑):
Mapping types:
MyIEnumerable`1 -> IMyEnumerable`1
MyIEnumerable`1[[DestinationItem]] -> IMyEnumerable`1[[DestinationItem]]
Destination path:
Destination.Items.Items
Source value:
MyIEnumerable`1[DestinationItem]
----> System.ArgumentException : Object of type System.Collections.Generic.List`1[DestinationItem] cannot be converted to type IMyEnumerable`1[DestinationItem].
我知道如何修复映射以便我可以使用它吗?
答案 0 :(得分:0)
假设如下:
var source = new Source
{
Items = new List<SourceItem>
{
new SourceItem { Name = "foo" },
new SourceItem { Name = "bar" },
new SourceItem { Name = "cow" },
}
};
接下来的工作:
// Method 1: Straight up mapping the collections:
Mapper.CreateMap<List<SourceItem>, IMyEnumerable<DestinationItem>>()
.ConstructUsing(list => new MyEnumerable<DestinationItem>(list.ConvertAll(Mapper.Map<SourceItem, DestinationItem>)));
// Method 2: Ignore the property and do it ourselves after the rest of the mapping:
Mapper.CreateMap<Source, Destination>()
.ForMember(q => q.Items, r => r.Ignore())
.AfterMap((s, d) => d.Items = new MyEnumerable<DestinationItem>(
s.Items.Select(Mapper.Map<SourceItem, DestinationItem>)));
由于List<T>
,IEnumerable<T>
和IMyEnumerable<T>