小组加入Linq C#

时间:2015-11-29 15:14:00

标签: c# linq

我正在阅读很多网站,以便更好地了解Linq -Group Join。

var customers = new Customer[]
{
    new Customer{Code = 5, Name = "Sam"},
    new Customer{Code = 6, Name = "Dave"},
    new Customer{Code = 7, Name = "Julia"},
    new Customer{Code = 8, Name = "Sue"}
};

// Example orders.
var orders = new Order[]
{
    new Order{KeyCode = 5, Product = "Book"},
    new Order{KeyCode = 6, Product = "Game"},
    new Order{KeyCode = 7, Product = "Computer"},
    new Order{KeyCode = 7, Product = "Mouse"},
    new Order{KeyCode = 8, Product = "Shirt"},
    new Order{KeyCode = 5, Product = "Underwear"}
};
        var query = customers.GroupJoin(orders,
          c => c.Code,
          o => o.KeyCode,
          (c, result) => new Result(c.Name, result));//why mention c here??

        // Enumerate results.
        foreach (var result in query)
        {
            Console.WriteLine("{0} bought...", result.Name);
            foreach (var item in result.Collection)
            {
                Console.WriteLine(item.Product);
            }
        }

我无法理解它为什么给出(c,结果)?如果写成(c,o)怎么办?

有人可以就此分享想法吗?

1 个答案:

答案 0 :(得分:2)

这些只是传递给Func的参数的名称。如果能让代码更清晰,您可以使用任何您想要的名称,即:

        var query = customers.GroupJoin(orders,
        c => c.Code,
        o => o.KeyCode,
        (something1, something2) => new Result(something1.Name, something2));

因为它只会将前两个Func中的参数传递给Func<TOuter, IEnumerable<TInner>, TResult>的最后一个,所以在这种情况下Func<Customer, IEnumerable<Order>, Result>

与这种情况相同:

public Result DoStuff(Order nameMeAnyWayYouWant, Customer meToo)
{
    //do stuff here
}

来自问题的代码来自:http://www.dotnetperls.com/groupjoin

我正在添加作者跳过的模型类,如果有人想详细说明,以及dotnetperls.com发生故障:

class Customer
{
    public int Code { get; set; }
    public string Name { get; set; }
}

class Order
{
    public int KeyCode { get; set; }
    public string Product { get; set; }
}

class Result
{
    public string Name { get; set; }
    public IEnumerable<Order> Collection { get; set; }
    public Result(string name, IEnumerable<Order> collection)
    {
        this.Name = name;
        his.Collection = collection;
    }
}