Automapper：如何利用自定义的INamingConvention？

Question

我正在设计一个数据库，设计师似乎很喜欢大写字母和下划线键。由于我有一个简单的ORM，我的数据模型也使用这些名称。我需要建立DTO，我宁愿给他们标准的名字，因为我们通过服务暴露他们。 下面的代码现在已更正！如果您需要使用多个命名约定

using System; 
    using System.Collections.Generic; 
    using System.Text.RegularExpressions; 
    using AutoMapper; 
    using NUnit.Framework; 

    namespace AutomapperTest 
    { 
     public class DATAMODEL 
     { 
      public Guid ID { get; set; } 
      public string FIRST_NAME { get; set; } 
      public List<CHILD_DATAMODEL> CHILDREN { get; set; } 
     } 

     public class CHILD_DATAMODEL 
     { 
      public Guid ID { get; set; } 
      public int ORDER_ID { get; set; } 
     } 

     public class DataModelDto 
     { 
      public Guid Id { get; set; } 
      public string FirstName { get; set; } 
      public List<ChildDataModelDto> Children { get; set; } 
     } 

     public class ChildDataModelDto 
     { 
      public Guid Id { get; set; } 
      public int OrderId { get; set; } 
     } 

     public class UpperUnderscoreNamingConvention : INamingConvention 
     { 
      private readonly Regex _splittingExpression = new Regex(@"[\p{Lu}0-9]+(?=_?)"); 

      public Regex SplittingExpression { get { return _splittingExpression; } } 

      public string SeparatorCharacter { get { return "_"; } } 
     } 

     public class Profile1 : Profile 
     { 
      protected override void Configure() 
      { 
       SourceMemberNamingConvention = new UpperUnderscoreNamingConvention(); 
       DestinationMemberNamingConvention = new PascalCaseNamingConvention(); 
       CreateMap<DATAMODEL, DataModelDto>(); 
       CreateMap<CHILD_DATAMODEL, ChildDataModelDto>(); 
      } 
     } 
     [TestFixture] 
     public class Tests 
     { 
      [Test] 
      public void CanMap() 
      { 
       //tell automapper to use my convention 
       Mapper.Initialize(x => x.AddProfile<Profile1>()); 
       //make a dummy source object 
       var src = new DATAMODEL(); 
       src.ID = Guid.NewGuid(); 
       src.FIRST_NAME = "foobar"; 
       src.CHILDREN = new List<CHILD_DATAMODEL> 
           { 
            new CHILD_DATAMODEL() 
             { 
              ID = Guid.NewGuid(), 
              ORDER_ID = 999 
             } 
           }; 
       //map to destination 
       var dest = Mapper.Map<DATAMODEL, DataModelDto>(src); 
       Assert.AreEqual(src.ID, dest.Id); 
       Assert.AreEqual(src.FIRST_NAME, dest.FirstName); 
       Assert.AreEqual(src.CHILDREN.Count, dest.Children.Count); 
       Assert.AreEqual(src.CHILDREN[0].ID, dest.Children[0].Id); 
       Assert.AreEqual(src.CHILDREN[0].ORDER_ID, dest.Children[0].OrderId); 
      } 
     } 
    } 

Answer 1

在配置文件中创建的映射，并定义INamingConvention参数适当的测试通过了所以用这个作为参考。

我不喜欢全局/静态，所以我更喜欢使用Initialize并将所有映射定义在一起。这还有一个额外的好处，就是允许调用AssertConfiguration ...这意味着如果我使用映射，我将在启动时得到异常，而不是每当我的代码开始使用有问题的映射时。

Mapper.Initialize(configuration => 
{ 
    configuration.CreateProfile("Profile1", CreateProfile1); 
    configuration.CreateProfile("Profile2", CreateProfile2); 
}); 
Mapper.AssertConfigurationIsValid(); 

在与初始化方法相同类

：

public void CreateProfile1(IProfileExpression profile) 
{ 
    // this.CreateMap (not Mapper.CreateMap) statements that do the "normal" thing here 
    // equivalent to Mapper.CreateMap(...).WithProfile("Profile1"); 
} 

public void CreateProfile2(IProfileExpression profile) 
{ 
    profile.SourceMemberNamingConvention = new PascalCaseNamingConvention(); 
    profile.DestinationMemberNamingConvention = new LowerUnderscoreNamingConvention(); 

    // this.CreateMap (not Mapper.CreateMap) statements that need your special conventions here 
    // equivalent to Mapper.CreateMap(...).WithProfile("Profile2"); 
} 

，如果你这样做了，并且不限定在两个配置文件相同的映射，我不认为你需要什么从原来的问题中“填补空白”，它应该已经设置为做正确的事情。

Answer 2

什么

public class DATAMODELProfile : Profile 
{ 
    protected override void Configure() 
    { 
     Mapper.CreateMap<DATAMODEL, DATAMODEL>(); 
     Mapper.CreateMap<DATAMODEL, SOMETHINGELSE>(); 
     Mapper.CreateMap<DATAMODEL, DataModelDto>() 
      .ForMember(dest => dest.Id, opt => opt.MapFrom(src => src.ID)) 
      .ForMember(dest => dest.FirstName, opt => opt.MapFrom(src => src.FIRST_NAME)) 
      .ForMember(dest => dest.ChildDataModels, opt => opt.MapFrom(src => src.CHILD_DATAMODELS)); 
    } 
}