只是为了另一种观点，可能会帮助一些函数式程序员:

选择是映射 SelectMany是绑定的(或flatMap的Scala/Kotlin人)

我理解SelectMany工作起来像一个连接快捷方式。

所以你可以:

var orders = customers
             .Where(c => c.CustomerName == "Acme")
             .SelectMany(c => c.Orders);

SelectMany()允许您以一种需要第二个Select()或循环的方式折叠多维序列。

更多细节请参见这篇博客文章。

下面是一个用于测试的初始化小集合的代码示例:

class Program
{
    static void Main(string[] args)
    {
        List<Order> orders = new List<Order>
        {
            new Order
            {
                OrderID = "orderID1",
                OrderLines = new List<OrderLine>
                {
                    new OrderLine
                    {
                        ProductSKU = "SKU1",
                        Quantity = 1
                    },
                    new OrderLine
                    {
                        ProductSKU = "SKU2",
                        Quantity = 2
                    },
                    new OrderLine
                    {
                        ProductSKU = "SKU3",
                        Quantity = 3
                    }
                }
            },
            new Order
            {
                OrderID = "orderID2",
                OrderLines = new List<OrderLine>
                {
                    new OrderLine
                    {
                        ProductSKU = "SKU4",
                        Quantity = 4
                    },
                    new OrderLine
                    {
                        ProductSKU = "SKU5",
                        Quantity = 5
                    }
                }
            }
        };

        //required result is the list of all SKUs in orders
        List<string> allSKUs = new List<string>();

        //With Select case 2 foreach loops are required
        var flattenedOrdersLinesSelectCase = orders.Select(o => o.OrderLines);
        foreach (var flattenedOrderLine in flattenedOrdersLinesSelectCase)
        {
            foreach (OrderLine orderLine in flattenedOrderLine)
            {
                allSKUs.Add(orderLine.ProductSKU);
            }
        }

        //With SelectMany case only one foreach loop is required
        allSKUs = new List<string>();
        var flattenedOrdersLinesSelectManyCase = orders.SelectMany(o => o.OrderLines);
        foreach (var flattenedOrderLine in flattenedOrdersLinesSelectManyCase)
        {
            allSKUs.Add(flattenedOrderLine.ProductSKU);
        }

       //If the required result is flattened list which has OrderID, ProductSKU and Quantity,
       //SelectMany with selector is very helpful to get the required result
       //and allows avoiding own For loops what according to my experience do code faster when
       // hundreds of thousands of data rows must be operated
        List<OrderLineForReport> ordersLinesForReport = (List<OrderLineForReport>)orders.SelectMany(o => o.OrderLines,
            (o, ol) => new OrderLineForReport
            {
                OrderID = o.OrderID,
                ProductSKU = ol.ProductSKU,
                Quantity = ol.Quantity
            }).ToList();
    }
}
class Order
{
    public string OrderID { get; set; }
    public List<OrderLine> OrderLines { get; set; }
}
class OrderLine
{
    public string ProductSKU { get; set; }
    public int Quantity { get; set; }
}
class OrderLineForReport
{
    public string OrderID { get; set; }
    public string ProductSKU { get; set; }
    public int Quantity { get; set; }
}

假设你有一组国家

var countries = new[] { "France", "Italy" };

如果你对国家执行Select，你将得到数组中的每个元素IEnumerable<T>

IEnumerable<string> selectQuery = countries.Select(country => country);

在上面的代码中，国家表示指向数组中每个国家的字符串。现在迭代selectQuery以获得国家:

foreach(var country in selectQuery)
    Console.WriteLine(country);

// output
//
// France
// Italy

如果你想打印每个国家的字符，你必须使用嵌套foreach

foreach (var country in selectQuery)
{
    foreach (var charOfCountry in country)
    {
        Console.Write(charOfCountry + ", ");
    }
}

// output

// F, r, a, n, c, e, I, t, a, l, y,

好的。现在尝试对国家执行SelectMany。这一次SelectMany获取每个国家作为字符串(和以前一样)，因为字符串类型是一个字符的集合，SelectMany尝试将每个国家分为它的组成部分(字符)，然后返回一个字符的集合IEnumerable<T>

IEnumerable<char> selectManyQuery = countries.SelectMany(country => country);

在上面的代码中，country表示一个字符串，该字符串引用数组中的每个国家，但返回值是每个国家的字符

实际上SelectMany喜欢在集合中获取两层，并将第二层平铺为IEnumerable<T>

现在遍历selectManyQuery以获得每个国家的字符:

foreach(var charOfCountry in selectManyQuery)
    Console.Write(charOfCountry + ", ");

// output

// F, r, a, n, c, e, I, t, a, l, y,

Some SelectMany可能不需要。以下2个查询给出相同的结果。

Customers.Where(c=>c.Name=="Tom").SelectMany(c=>c.Orders)

Orders.Where(o=>o.Customer.Name=="Tom")

对于一对多关系，

如果从“1”开始，则需要SelectMany，它将平展many。 如果从“Many”开始，则不需要SelectMany。(仍然能够从“1”中过滤，这也比以下标准的连接查询更简单)

from o in Orders
join c in Customers on o.CustomerID equals c.ID
where c.Name == "Tom"
select o