演员/转换IEnumerable的< T>到了IEnumerable< U&GT ;? [英] Cast/Convert IEnumerable<T> to IEnumerable<U>?

问题描述

以下符合规定，但在运行时会抛出异常。我所试图做的是投类PersonWithAge一类人的。我如何做到这一点，什么是身边的工作？

The following complies but at run time throws an exception. What I am trying to do is to cast a class PersonWithAge to a class of Person. How do I do this and what is the work around?

class Person
{
	public int Id { get; set; }
	public string Name { get; set; }
}

class PersonWithAge
{
	public int Id { get; set; }
	public string Name { get; set; }
	public int Age { get; set; }
}

class Program
{
	static void Main(string[] args)
	{
		IEnumerable<PersonWithAge> pwa = new List<PersonWithAge>
        {
        	new PersonWithAge {Id = 1, Name = "name1", Age = 23},
        	new PersonWithAge {Id = 2, Name = "name2", Age = 32}
        };

		IEnumerable<Person> p = pwa.Cast<Person>();

		foreach (var i in p)
		{
			Console.WriteLine(i.Name);
		}
	}
}

编辑：顺便说一句PersonWithAge总是包含相同的属性，另加一对夫妇更。

By the way PersonWithAge will always contain the same properties as Person plus a couple more.

编辑2 对不起球员，但我应该有使这个更清楚一点，说我有一个包含相同的列，但视图2包含1额外字段数据库2分贝意见。我的模型视图实体是由模仿数据库视图工具生成的。我有一个从实体类的一个继承了MVC的部分观点，但我必须抓住的数据不止一种方法...

EDIT 2 Sorry guys but I should have made this a bit clearer, say I have two db views in a database that contains the same columns but view 2 contains 1 extra field. My model view entities are generated by a tool that mimics the database views. I have a MVC partial view that inherits from one of the class entities but I have more than one way to grab data...

不知道这是否帮助，但它意味着我不能让personWithAge由人继承。

Not sure if this helps but it means that I cant make personWithAge inherit from person.

推荐答案

您可以不投，因为它们是不同的类型。你有两个选择：

You can't cast because they are different types. You have two choices:

1）改变类，这样PersonWithAge从人继承

1) Change the class so that PersonWithAge inherits from person.

class PersonWithAge : Person
{
        public int Age { get; set; }
}

2）创建新的对象：

2) Create new objects:

IEnumerable<Person> p = pwa.Select(p => new Person { Id = p.Id, Name = p.Name });

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