SelectMany有几个重载。其中之一允许您在遍历层次结构时跟踪父节点和子节点之间的任何关系。

示例:假设您有以下结构:League -> Teams -> Player。

您可以很容易地返回一个平坦的播放器集合。但是你可能会失去任何关于玩家所在团队的信息。

幸运的是，有一个重载用于此目的:

var teamsAndTheirLeagues = 
         from helper in leagues.SelectMany
               ( l => l.Teams
                 , ( league, team ) => new { league, team } )
                      where helper.team.Players.Count > 2 
                           && helper.league.Teams.Count < 10
                           select new 
                                  { LeagueID = helper.league.ID
                                    , Team = helper.team 
                                   };

前面的例子来自Dan的IK博客。我强烈建议你看一看。

SelectMany扁平查询，返回列表的列表。例如

public class PhoneNumber
{
    public string Number { get; set; }
}

public class Person
{
    public IEnumerable<PhoneNumber> PhoneNumbers { get; set; }
    public string Name { get; set; }
}

IEnumerable<Person> people = new List<Person>();

// Select gets a list of lists of phone numbers
IEnumerable<IEnumerable<PhoneNumber>> phoneLists = people.Select(p => p.PhoneNumbers);

// SelectMany flattens it to just a list of phone numbers.
IEnumerable<PhoneNumber> phoneNumbers = people.SelectMany(p => p.PhoneNumbers);

// And to include data from the parent in the result: 
// pass an expression to the second parameter (resultSelector) in the overload:
var directory = people
   .SelectMany(p => p.PhoneNumbers,
               (parent, child) => new { parent.Name, child.Number });

Live Demo。net Fiddle

没有太多的技术-数据库与许多组织，每个有许多用户:-

var orgId = "123456789";

var userList1 = db.Organizations
                   .Where(a => a.OrganizationId == orgId)
                   .SelectMany(a => a.Users)
                   .ToList();

var userList2 = db.Users
                   .Where(a => a.OrganizationId == orgId)
                   .ToList();

都为所选组织返回相同的ApplicationUser列表。

第一个“项目”从Organization到Users，第二个直接查询Users表。

var players = db.SoccerTeams.Where(c => c.Country == "Spain")
                            .SelectMany(c => c.players);

foreach(var player in players)
{
    Console.WriteLine(player.LastName);
}

De Gea 阿尔芭 科斯塔 别墅 Busquets

...

SelectMany()方法用于抚平序列，其中序列的每个元素都是独立的。

我的类用户是这样的

class User
    {
        public string UserName { get; set; }
        public List<string> Roles { get; set; }
    }

主要:

var users = new List<User>
            {
                new User { UserName = "Reza" , Roles = new List<string>{"Superadmin" } },
                new User { UserName = "Amin" , Roles = new List<string>{"Guest","Reseption" } },
                new User { UserName = "Nima" , Roles = new List<string>{"Nurse","Guest" } },
            };

var query = users.SelectMany(user => user.Roles, (user, role) => new { user.UserName, role });

foreach (var obj in query)
{
    Console.WriteLine(obj);
}
//output

//{ UserName = Reza, role = Superadmin }
//{ UserName = Amin, role = Guest }
//{ UserName = Amin, role = Reseption }
//{ UserName = Nima, role = Nurse }
//{ UserName = Nima, role = Guest }

您可以对序列的任何项使用操作

int[][] numbers = {
                new[] {1, 2, 3},
                new[] {4},
                new[] {5, 6 , 6 , 2 , 7, 8},
                new[] {12, 14}
            };

IEnumerable<int> result = numbers
                .SelectMany(array => array.Distinct())
                .OrderBy(x => x);

//output

//{ 1, 2 , 2 , 3, 4, 5, 6, 7, 8, 12, 14 }

 List<List<int>> numbers = new List<List<int>> {
                new List<int> {1, 2, 3},
                new List<int> {12},
                new List<int> {5, 6, 5, 7},
                new List<int> {10, 10, 10, 12}
            };

 IEnumerable<int> result = numbers
                .SelectMany(list => list)
                .Distinct()
                .OrderBy(x=>x);

//output

// { 1, 2, 3, 5, 6, 7, 10, 12 }

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