6

I have populated the follow list with objects of type AnonymousType

List<object> someList = new List<object>();

someList.Add(new { foo = 1 });

My problem is that I can't make it stronly typed to do something like this:

someList.Where(x=> x.foo == 1);

However, it's possible on this list:

var someList = new[] { new { foo = 1 } };

Can I cast my first list to make it behave like the second list? I want to be able to use lambda expressions on the properties like I showed above.


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4

2 回答 2

9

您可以利用泛型和类型推断为您创建列表:

public static List<T> CreateAnonymousList<T>(params T[] entries)
{
    return new List<T>(entries);
}

用法如:

var someList = CreateAnonymousList(new { foo = 1 }, new { foo = 2 }, new { foo = 1 });

someList.Where(x => x.foo == 1);

自然,您将无能为力您将永远无法在您的代码中将它强类型为任何东西,var或者从您的方法中返回它或者您通常无法使用匿名类型做的任何事情。如果你想做更多,你只需要咬紧牙关,为你的匿名类型定义一个类。


重新阅读您的问题,您仍然可以对数组执行 LINQ 查询:

var someArray = new[]{new { foo = 1 }, new { foo = 2 }, new { foo = 1 }};
someArray.Where(x => x.foo == 1)

因此,除非您正在修改它(例如通过像or之类的标准List<T>操作),否则没有理由将其转换为.AddRemoveList<T>

我意识到也许您仍然希望能够将其传回(出于某种原因)并且仍然在不知道其匿名类型的情况下对其进行操作。在这种情况下,您可以将其视为dynamic并在运行时执行操作,但您会丢失通常使用匿名类型的任何智能感知/强类型:

List<dynamic> someDynamicList = new List<dynamic>() {new { foo = 1 }, new { foo = 2 }, new { foo = 1 }};
someDynamicList.Where(x => x.foo == 1)

Tim Schmelter利用 Jon Skeet 的 CastByExample指出的最后一种方法,但扩展为使用扩展方法转换您的集合:

public static IEnumerable<T> CastByExample<T>(this IEnumerable source, T example)
{
    foreach(object entry in source)
        yield return (T)entry;
}

public static IEnumerable CreateAnonymousData()
{
    return new[]{new { foo = 1 }, new { foo = 2 }, new { foo = 1 }};
}

使用如下:

var anonymousData = CreateAnonymousData();
var typedAnonymousData = anonymousData.CastByExample(new { foo = 1 });
typedAnonymousData.Where(x => x.foo == 1);

这利用了这样一个事实,即在同一个程序集中,使用相同参数名称、类型和顺序声明的匿名类型编译为相同类型。如果您需要CreateAnonymousData从当前程序集外部调用您的,并且您必须在使用它的任何地方维护您的foo匿名类型的签名(添加/更改其签名,您必须在使用它的任何地方更新它,或者您正在会过得很糟糕)。

但我认为现在越来越清楚,最好的解决方案是简单地定义一个匿名类型的类表示。

于 2013-07-05T12:41:02.427 回答
4

您可以使用 Jon Skeets' CastByExample

public static T CastByExample<T>(object input, T example)
{
    return (T)input;
}

List<object> someList = new List<object>() { 
    new { foo = 1 },new { foo = 2 },new { foo = 3 }
};

var example = new { foo = 0 };

foreach (object obj in someList)
{
    var x = CastByExample(obj, example);
    Console.WriteLine("Foo: " + x.foo);
}
于 2013-07-05T12:48:58.687 回答