How to use AutoMapper to map destination object with a child object in the source object?

匿名 (未验证) 提交于 2019-12-03 01:11:01

问题:

I have the source and destination objects like this:

class ProductWithCategories // Source class {     public Product Product { get; set; } // Product is an EF entity class     public IEnumerable<Category> Categories { get; set; } }  class ProductViewModel // Dest class {     public int Id { get; set; }     // Other properties with the same name as Product class      public IEnumerable<CategoryViewModel> Categories { get; set; } } 

So, my need is to map the values of source.Product into dest, and then source.Categories into dest.Categories. Is it possible with AutoMapper?

I have tried this and I was not surprised when it failed:

        config.CreateMap<ProductWithCategories, ProductViewModel>()             .ForMember(q => q, option => option.MapFrom(q => q.Product))             .ForMember(q => q.Categories, option => option.MapFrom(q => q.Categories)); 

Here is the exception I received:

[AutoMapperConfigurationException: Custom configuration for members is only supported for top-level individual members on a type.]

回答1:

After some discussion with OP, it turns out his main need is to quickly map a child/nested object inside the source object to the flattened destination object. He does not want to write a mapping for every property of the destination.

Here is a way to achieve this:

  • Define a mapping Product -> ProductViewModel used to flatten the members of Product
  • Define a mapping Category to CategoryViewModel
  • Define a mapping ProductWithCategories -> ProductViewModel that maps the categories, and then in the aftermap, map the Product:

    config.CreateMap<ProductWithCategories, ProductViewModel>() .ForMember(q => q.Id, option => option.Ignore()) // flattened in AfterMap .ForMember(q => q.Categories, option => option.MapFrom(q => q.Categories)) .AfterMap((src, dst) => Mapper.Map(src.Product, dst));



回答2:

Using recent versions of AutoMapper, you can do something like the following:

config.CreateMap<Product, ProductViewModel>()       .ForMember(q => q.Categories, option => option.Ignore());  config.CreateMap<ProductWithCategories, ProductViewModel>()       .ConstructUsing(s => AutoMapper.Mapper.Map<ProductViewModel>(s.Product))       .ForMember(q => q.Categories, option => option.MapFrom(q => q.Categories))       .ForAllOtherMembers(o => o.Ignore(); 

ConstructUsing() is used to generate and populate the base class from the nested child[ren] of the source. If you have more than one such nested child, you would need to make several mapping calls to map each of them successively onto the instance generated by the first Map() call. The .ForAllOtherMembers() is relatively recent (if you don't have it, get a newer version of AutoMapper.) Unfortunately it's slightly unsafe as if you add destination members which will need mapping but forget to update the map, configuration validation will not catch it.



回答3:

The offending line that generates the error is

.ForMember(q => q, option => option.MapFrom(q => q.Product)) 

The error message is hard to understand, but it means you have to state the destination properties explicitly:

.ForMember(q => q.Id, option => option.MapFrom(q => q.Product.Id)) .ForMember(q => q.OtherProperty, option => option.MapFrom(q => q.Product.OtherProperty)) 

You also have to define a mapping from Category to CategoryViewModel for

.ForMember(q => q.Categories, option => option.MapFrom(q => q.Categories)) 

to work:

config.CreateMap<Category, CategoryViewModel>(); 


回答4:

you should do like -

AutoMapper.Mapper.CreateMap<Category, CategoryViewModel>(); AutoMapper.Mapper.CreateMap<ProductWithCategories, ProductViewModel>()      .ForMember(a => a.Id, b => b.ResolveUsing(c => c.Product != null ? c.Product.MyProperty : 0))      .ForMember(a => a.Categories, b => b.ResolveUsing(c => c.Categories)); 

But it is better to wrap those properties from ProductViewModel (props like Id) inside another class. And create another map for things to work automapper way.



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