[C#] 使用 AutoMapper 快速轉換有相同資料的 Model

因為工作因素發生了需要將兩個具有相同資料的Model進行轉換，但是如果手動轉換的話太累了，工程師就是懶嘛~
於是乎就拜了Google大神找到了一個名叫AutoMapper套件。

1. 從NuGet下載AutoMapper
   
2. 新增兩組相同的Model

   class Model_1
   {
   	public string Name { get; set; }
   
   	public int Year { get; set; }
   
   	public DateTime Date { get; set; }
   }
   
   class Model_2
   {
   	public string Name { get; set; }
   
   	public int Year { get; set; }
   
   	public DateTime Date { get; set; }
   }
   

3. 轉換用的方法

   static TModel_2 ConvertModelToModel<TModel_1, TModel_2>(TModel_1 list)
   {
   	Mapper.Initialize(cfg =>
   	{
   		cfg.CreateMissingTypeMaps = true;
   		cfg.CreateMap<TModel_1, TModel_2>().ReverseMap();
   	});
   	Mapper.Configuration.AssertConfigurationIsValid();
   
   	var converted = Mapper.Map<TModel_2>(list);
   	return converted;
   }

4. 轉換

   static void Main(string[] args)
   {
   	Model_1 model1 = new Model_1();
   	Model_2 model2 = new Model_2();
   
   
   	model1.Name = "王小明";
   	model1.Year = 2016;
   	model1.Date = DateTime.UtcNow.AddHours(8);
   
   
   	Console.WriteLine("=== 轉換前 ===");
   	Console.WriteLine("model1=name:{0}, year:{1}, date:{2}, type:{3}", model1.Name, model1.Year, model1.Date, model1.GetType());
   	Console.WriteLine("model2=name:{0}, year:{1}, date:{2}, type:{3}", model2.Name, model2.Year, model2.Date, model2.GetType());
   
   	model2 = ConvertModelToModel<Model_1, Model_2>(model1);
   
   	Console.WriteLine("=== 轉換後 ===");
   	Console.WriteLine("model1=name:{0}, year:{1}, date:{2}, type:{3}", model1.Name, model1.Year, model1.Date, model1.GetType());
   	Console.WriteLine("model2=name:{0}, year:{1}, date:{2}, type:{3}", model2.Name, model2.Year, model2.Date, model2.GetType());
   
   	Console.ReadLine();
   }

5. 結果
   

原始碼：https://github.com/shuangrain/ConsoleApplication_AutoMapper