如何實現自定義命名解析爲AutoMapper

Question

如何映射使用AutoMapper 下面的類沒有明確地表明所有成員映射：使用翻譯類

public class Destination 
{ 
    public string Imie { get; set; } 
    public string Nazwisko { get; set; } 
    ... (huge amount of other properties) 
} 

：

public class Source 
{ 
    public string FirstName { get; set; } 
    public string LastName { get; set; } 
    ... (huge amount of other properties) 
} 

類

var translations = new Dictionary<string, string>() 
{ 
    { "FirstName", "Imie" }, 
    { "LastName", "Nazwisko" }, 
    ... (huge amount of other translations) 
} 

Answer 1

一種方法是使用Reflection：

private TDestination Map<TSource, TDestination>(TSource source, Dictionary<string, string> mapData) 
     where TSource : class 
     where TDestination : class, new() 
{ 
    if (source == null) return null; 
    if (mapData == null) mapData = new Dictionary<string, string>(); 

    TDestination destination = new TDestination(); 
    PropertyInfo[] sourceProperties = typeof(TSource).GetProperties(); 
    foreach (PropertyInfo property in sourceProperties) 
    { 
     string destPropertyName = mapData.ContainsKey(property.Name) ? mapData[property.Name] : property.Name; 
     PropertyInfo destProperty = typeof(TDestination).GetProperty(destPropertyName); 
     if (destProperty == null) continue; 
     destProperty.SetValue(destination, property.GetValue(source)); 
    } 
    return destination; 
} 

你會這樣稱呼它：

Destination mapped = Map<Source, Destination>(source, translations); 

Answer 2

這裏是你可以怎麼做。

考慮以下方法：

public void CreateMapBasedOnDictionary<TSource, TDestination>(IDictionary<string, string> mapping_dictionary) 
{ 
    var mapping_expression = AutoMapper.Mapper.CreateMap<TSource, TDestination>(); 

    foreach (var kvp in mapping_dictionary) 
    { 
     string source_property_name = kvp.Key; 
     string destination_property_name = kvp.Value; 

     Type member_type = typeof (TSource).GetProperty(source_property_name).PropertyType; 

     mapping_expression = mapping_expression.ForMember(destination_property_name, x => 
     { 
      typeof (IMemberConfigurationExpression<TSource>) 
       .GetMethod("MapFrom", new []{typeof(string)}) 
       .MakeGenericMethod(member_type) 
       .Invoke(x, new[] { source_property_name }); 
     }); 
    } 
} 

然後你就可以使用它像這樣：

var translations = new Dictionary<string, string>() 
{ 
    {"FirstName", "Imie"}, 
    {"LastName", "Nazwisko"}, 

}; 

CreateMapBasedOnDictionary<Source, Destination>(translations); 

Source src = new Source() 
{ 
    FirstName = "My first name", 
    LastName = "My last name" 
}; 

var dst = AutoMapper.Mapper.Map<Destination>(src); 

下面是CreateMapBasedOnDictionary方法的解釋：

AutoMapper已經有過載ForMember，允許您按名稱指定目的地屬性。我們在這裏很好。

它也有MapFrom的重載，允許您按名稱指定源屬性。然而，這種超載的問題在於它需要一個通用參數（TMember）作爲屬性類型。

我們可以通過使用反射來獲取屬性的類型，然後動態調用MapFrom方法和適當的TMember類型參數來解決此問題。

Answer 3

感謝@雅各布 - 馬薩德答案，我有此解決方案：

var map = Mapper.CreateMap<Source, Destination>(); 

var sourceProperties = typeof(Source).GetProperties(); 
foreach (var sourceProperty in sourceProperties) 
{ 
    var sourcePropertyName = sourceProperty.Name; 
    var destinationPropertyName = translations[sourceProperty.Name]; 

    map.ForMember(destinationPropertyName, mo => mo.MapFrom<string>(sourcePropertyName)); 
} 

var src = new Source() { FirstName = "Maciej", LastName = "Lis" }; 
var dest = Mapper.DynamicMap<Destination>(src);