Linq SelectMany 交叉连接

 

示例一

    internal class Program
    {
        private static void Main(string[] args)
        {
            List<Teacher> teachers = new List<Teacher>
            {
                new Teacher("a",new List<Student1>{ new Student1(),new Student1(),new Student1() }),
                new Teacher("b",new List<Student1>{ new Student1(),new Student1(),new Student1() }),
                new Teacher("c",new List<Student1>{ new Student1(),new Student1(),new Student1() }),
                new Teacher("d",new List<Student1>{ new Student1(),new Student1(),new Student1() }),
                new Teacher("e",new List<Student1>{ new Student1(),new Student1(),new Student1() }),
                new Teacher("f",new List<Student1>{ new Student1(),new Student1(),new Student1() }),
                new Teacher("g",new List<Student1>{ new Student1(),new Student1(),new Student1() })
            };

            IEnumerable<Student1> t1 = teachers.SelectMany(s => s.Students.Where(w => w.Score < ));
            t1.ToList().ForEach(f => Console.WriteLine(f));

            Console.Read();
        }
    }

    internal class Student1
    {
        public int Score { get; set; }

        public Student1(int score)
        {
            Score = score;
        }

        public override string ToString()
        {
            return $"studen1 : {Score}";
        }
    }

    internal class Teacher
    {
        public string Name { get; set; }

        public List<Student1> Students;

        public Teacher(string order, List<Student1> students)
        {
            Name = order;
            Students = students;
        }
    }

示例二

                    var query = from left in Enumerable.Range(, )
                                from right in Enumerable.Range(, left)
                                select new { left, right };

                    var result = Enumerable.Range(, ).SelectMany(left => Enumerable.Range(, left), (left, right) => new { left, right });

                    foreach (var item in result)
                    {
                        Console.WriteLine(item);
                    }