Best way to create a map between two entities with a third one from another context - .net-core

Question

Hi I'd like to create a map between two entities (source: User, target: UserInfosDto) while one member of the target DTO (UserItemPreference) needs info from a third entity inside another context.
public class UserInfosDto
{
//public int UserId { get; set; }
public string FirstName { get; set; }
public string LastName { get; set; }
public string Email { get; set; }
public UserItemPreferencesDto UserItemPreferences { get; set; }
}
public class UserItemPreferencesDto
{
public bool SeeActuality { get; set; }
public bool IsInEditorMode { get; set; }
}
public class User
{
public string IdentityId { get; set; }
//...
public string FirstName { get; set; }
public string LastName { get; set; }
public string Email { get; set; }
}
public class UserIdentity
{
public string IdentityId { get; set; }
//...
public bool SeeActuality { get; set; }
public bool IsInEditorMode { get; set; }
}
User and UserIdentity come from different databases but have a common property IdentityId. I thought about using ITypeConverter in which I would inject the UserIdentity dbContext. Problem is that I can't find a way to use ITypeConverter on one member only.

Answer 1

Use an IValueResolver instead, which allows to resolve separate members instead of full types.
For your case above it will look like
public class UserItemPreferencesResolver
: IValueResolver<User, UserInfosDto, UserItemPreferencesDto>
{
private readonly UserEntityDbContext _dbContext;
public UserItemPreferencesResolver(UserEntityDbContext dbContext)
{
_dbContext = dbContext;
}
public UserItemPreferencesDto Resolve(
User source,
UserInfosDto destination,
UserItemPreferencesDto destinationMember,
ResolutionContext context
)
{
UserItemPreferencesDto preferences = /* resolve from _dbContext (and transform) */
return preferences;
}
}
Your create the mapping via
CreateMap<User, UserInfosDto>()
.ForMember(
dest => dest.UserItemPreferences,
opt => opt.MapFrom<UserItemPreferencesResolver>()
);