C# 自动映射:从自动映射对象解析源属性名称
鉴于以下类别:C# 自动映射:从自动映射对象解析源属性名称,c#,.net,automapper,C#,.net,Automapper,鉴于以下类别: public class User { public int Id {get;set;} public PersonName Name {get;set;} } public class PersonName { public string FirstName {get;set;} public string LastName {get;set;} } public class UserDto { public int Id {get;set;}
public class User
{
public int Id {get;set;}
public PersonName Name {get;set;}
}
public class PersonName
{
public string FirstName {get;set;}
public string LastName {get;set;}
}
public class UserDto
{
public int Id {get;set;}
public string FirstName {get;set;}
}
以及以下映射配置:
Mapper.CreateMap<User, UserDto>()
.ForMember(destination => destination.FirstName,
options => options.MapFrom(source => source.Name.FirstName))
Mapper.CreateMap()
.ForMember(destination=>destination.FirstName,
options=>options.MapFrom(source=>source.Name.FirstName))
是否可以解析目标对象上给定属性的源属性名称:
比如:
Assert.AreEqual(GetSourcePropertyName<User, UserDto>("FirstName"), "Name.FirstName")
Assert.AreEqual(GetSourcePropertyName(“FirstName”),“Name.FirstName”)
因为MapFrom()接受lambda,所以目标属性可能映射到任何对象。
你可以使用任何你想要的lambda。考虑这一点:
.ForMember(
destination => destination.FullName,
options => options.MapFrom(source => source.Name.FirstName + " " + source.Name.LastName)
);
因为不必强制创建简单属性访问器lambdas,所以不能将源表达式简化为简单属性名称字符串
如果MapFrom()采用Expression
,则可以将表达式转换为字符串,但无法按当前编写的方式执行