94

I have the classes:

public class Person{ /* Props here */ }

public class PersonViewModel { /* Props here */ }

Then the list:

List<Person> people = new List<Person>();
List<PersonViewModel> peopleVM = Mapper
                                .MapList<Person, PersonViewModel>(people); //Problem here.

What is the correct way to do this?

6 Answers 6

116
Mapper.CreateMap<Person, PersonViewModel>();
peopleVM = Mapper.Map<List<Person>, List<PersonViewModel>>(people);
Mapper.AssertConfigurationIsValid();

From Getting Started:

How do I use AutoMapper?

First, you need both a source and destination type to work with. The destination type's design can be influenced by the layer in which it lives, but AutoMapper works best as long as the names of the members match up to the source type's members. If you have a source member called "FirstName", this will automatically be mapped to a destination member with the name "FirstName". AutoMapper also supports Flattening, which can get rid of all those pesky null reference exceptions you might encounter along the way.

Once you have your types, and a reference to AutoMapper, you can create a map for the two types.

Mapper.CreateMap<Order, OrderDto>();

The type on the left is the source type, and the type on the right is the destination type. To perform a mapping, use the Map method.

OrderDto dto = Mapper.Map<Order, OrderDto>(order);
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12 Comments

Do I need to use CreateMap? I've been using automapper without it and it works.
Probably not as long as property names match and it doesn't need to convert anything. You can use create map if property names don't match or you want to ignore properties etc.
Try adding the create map. You can also do Mapper.AssertConfigurationIsValid(); after the create. Usually you put all your creates in bootstrapper or startup function. AutoMapper also supports profiles.
Is there any short method to create mapping to both direction to each other? Mapper.Map<List<Person>, List<PersonViewModel>>(person) and Mapper.Map<List<PersonViewModel> ,List<Person>>(person)
@doganak Mapper.Map<List<Person>, List<PersonViewModel>>(person).ReverseMap()
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30

Another Solution

List<Person> people = new List<Person>();
List<PersonViewModel> peopelVM;
peopelVM = people.Select(Mapper.Map<Person, PersonViewModel>);

And in the Automapper config

Mapper.CreateMap<Person, PersonViewModel>();

Comments

15

If you're using IQueryable lists here (from EF or NH, for example) you can use the AutoMapper.IQueryableExtensions methods, Project() and To().

This is my first time with AutoMapper, but I'm succeeding by creating a map for just the model:

Mapper.CreateMap<Person, PersonViewModel>();
Mapper.AssertConfigurationIsValid();

And then using the IQueryableExtension methods Project() and To():

using AutoMapper.QueryableExtensions;
...

IQueryable<Person> people = new List<Person>().AsQueryable(); //actually from ORM
IQueryable<PersonViewModel> peopleVM = people.Project().To<PersonViewModel>();

1 Comment

I find this answer the most useful / elegant. However, Project() is now deprecated and should be replaced with ProjectTo().
3

In core 1.1 this extension might work:

public static List<TDestination> MapList<TSource, TDestination>(this IMapper mapper, List<TSource> source)
        {
            return source.Select(x => mapper.Map<TDestination>(x)).ToList();
        }

Comments

3

Another Solution

mapper.Map<IEnumerable<PersonViewModel>>(people);

Comments

0

You could create an extension method to do something like this using existing mappings for individual items:

public static class AutoMapperExtensions
{
    public static List<TDestination> MapList<TSource, TDestination>(this IMapper mapper, List<TSource> source)
    {
        return mapper.Map<List<TDestination>>(source);
    }
}

Usage:

List<PersonViewModel> peopleVM = _mapper.MapList<PersonViewModel>(people);

Comments

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