lambda selectMany （就是将查询出来的多个集合，合并到一起，返回一个合并后的集合）

class Person
    {
        public string Name { set; get; }
        public int Age { set; get; }
        public string Gender { set; get; }
        public Dog[] Dogs { set; get; }
    }
public class Dog
    {
        public string Name { set; get; }
    }

List<Person> personList = new List<Person>
            {
                new Person
                {
                    Name = "P1", Age = 18, Gender = "Male",
                    Gogs = new Dog[]
                    {
                        new Dog { Name = "D1" },
                        new Dog { Name = "D2" }
                    }
                },
                new Person
                {
                    Name = "P2", Age = 19, Gender = "Male",
                    Gogs = new Dog[]
                    {
                        new Dog { Name = "D3" }
                    }
                },
                new Person
                {
                    Name = "P3", Age = 17,Gender = "Female",
                    Dogs = new Dog[]
                    {
                        new Dog { Name = "D4" },
                        new Dog { Name = "D5" },
                        new Dog { Name = "D6" }
                    }
                }
            };

我们可以看到p1对应D1,D2

p2对应D3            p3对应D4D5D6

最基本用法

var dogs = personList.SelectMany(p => p.Dogs);
            foreach (var dog in dogs)
            {
                Console.WriteLine(dog.Name);
            }

重载用法

var dogs = personList.SelectMany((p, i) => 
    p.Dogs.Select( d=>
    {
        d.Name = $"{i},{d.Name}";
        return d;
    }));

 

重载用法

var results = personList.SelectMany(p => p.Dogs, (p, d) => new { PersonName = p.Name, DogName = d.Name });
            foreach (var result in results)
            {
                Console.WriteLine($"{result.PersonName},{result.DogName}");
            }
// p就是前面的p，也就是迭代对象
d是p.dogs 里面的元素，也就是dog

第二个参数类似于SQL的交叉集合

 最后一种重载

var results = personList.SelectMany((p,i) =>
{
    for(int j=0;j<p.Dogs.Length;j++)
    {
        p.Dogs[j].Name = $"{i}-{p.Dogs[j].Name}";
    }
    return p.Dogs;
}, (p, d) => new { PersonName = p.Name, DogName = d.Name });