CODEWITHSUNDEEP
c#dependency-injectionautomapper
Tom
Tom

Reputation: 1581

Using dependency injection with AutoMapper ValueResolver that has constructor arguments

I have the following custom value resolver:

public class ImageUrlResolver<T> : ValueResolver<T, string>
{
    private readonly ISettings _settings;

    public ImageUrlResolver(string size)
    {
        _settings = ObjectFactory.GetInstance<ISettings>();
    }

    ...
}

.ForMember(d => d.ImageUrl, 
    o => o.ResolveUsing<ImageUrlResolver>().ConstructedBy(() => new ImageUrlResolver("150x150"))

I'm trying to update it so that I can inject StructureMap's IContainer instead of using ObjectFactory, but I'm not sure how I can construct the resolver when it has constructor arguments. Is there anything else I can do?

Upvotes: 3

Views: 4369

Answers (1)

Tom
Tom

Reputation: 1581

I've figured out a solution. I'm now injecting IContainer into the profile, and passing it through to the resolver.

public static void Initialise(IContainer container)
{
    var type = typeof(Profile);
    var profiles = AppDomain.CurrentDomain
                            .GetAssemblies()
                            .SelectMany(a => a.GetTypes())
                            .Where(t => type.IsAssignableFrom(t) && type != t)
                            .Select(container.GetInstance)
                            .Cast<Profile>()
                            .ToList();

    Mapper.Initialize(c =>
        {
            profiles.ForEach(c.AddProfile);
            c.ConstructServicesUsing(container.GetInstance);
        });
}

public class MyProfile : Profile
{
    private readonly IContainer _container;

    public MyProfile(IContainer container)
    {
        _container = container;
    }

    private static void Configure()
    {
        Mapper.CreateMap<Entity, Model>()
            .ForMember(d => d.ImageUrl, o => o.ResolveUsing<ImageUrlResolver>().ConstructedBy(() => new ImageUrlResolver(_container, "150x150"))
    }
}

Maybe not the cleanest solution, but it's the only one I found that works.

Upvotes: 1

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