您可以使用反射 http://msdn.microsoft.com/en-us/library/ms173183(VS.80).aspx为了这。
您的场景可能看起来有点像这样:
static void Main(string[] args)
{
var list = new List<Mammal>();
list.Add(new Person { Name = "Filip", DOB = DateTime.Now });
list.Add(new Person { Name = "Peter", DOB = DateTime.Now });
list.Add(new Person { Name = "Goran", DOB = DateTime.Now });
list.Add(new Person { Name = "Markus", DOB = DateTime.Now });
list.Add(new Dog { Name = "Sparky", Breed = "Unknown" });
list.Add(new Dog { Name = "Little Kid", Breed = "Unknown" });
list.Add(new Dog { Name = "Zorro", Breed = "Unknown" });
foreach (var item in list)
Console.WriteLine(item.Speek());
list = ReCalculateDOB(list);
foreach (var item in list)
Console.WriteLine(item.Speek());
}
您想要重新计算所有哺乳动物的生日的地方。上面的实现看起来像这样:
internal interface Mammal
{
string Speek();
}
internal class Person : Mammal
{
public string Name { get; set; }
public DateTime DOB { get; set; }
public string Speek()
{
return "My DOB is: " + DOB.ToString() ;
}
}
internal class Dog : Mammal
{
public string Name { get; set; }
public string Breed { get; set; }
public string Speek()
{
return "Woff!";
}
}
因此,基本上您需要做的是使用 Relfection,这是一种在运行时检查类型并获取类型属性和其他类似内容的机制。以下是如何为每只获得出生日期的哺乳动物在上述出生日期基础上添加 10 天的示例。
static List<Mammal> ReCalculateDOB(List<Mammal> list)
{
foreach (var item in list)
{
var properties = item.GetType().GetProperties();
foreach (var property in properties)
{
if (property.PropertyType == typeof(DateTime))
property.SetValue(item, ((DateTime)property.GetValue(item, null)).AddDays(10), null);
}
}
return list;
}
请记住,使用反射可能会很慢,而且通常很慢。
但是,上面将打印:
My DOB is: 2010-03-22 09:18:12
My DOB is: 2010-03-22 09:18:12
My DOB is: 2010-03-22 09:18:12
My DOB is: 2010-03-22 09:18:12
Woff!
Woff!
Woff!
My DOB is: 2010-04-01 09:18:12
My DOB is: 2010-04-01 09:18:12
My DOB is: 2010-04-01 09:18:12
My DOB is: 2010-04-01 09:18:12
Woff!
Woff!
Woff!