IQueryable 提供对数据类型已知的特定数据源的查询计算的功能

/*1.使用Enumerable.Aggregate<TSource, TAccumulate, TResult> 方法： 对序列应用累加函数,将指定的种子作为初始值,并且使用指定的函数选择结构值*/

 string[] fruits= { "APPLE","banana","Pear","PEACH","watermelon" };

//初始值为Apple, 

//指定的函数用Lambda //

但是我这里有个疑问,这样选择之后只能有一个值，其实banana长度其实也是大于apple的,但是它的原理是用func结果替换以前的结果，要是能够返回结果集就好啦！

 string result= fruits.Aggregate

                                   ("APPLE" ,

                                   (start, end) => end.Length > start.Length ? start = end : start );

Console.WriteLine(result);

 

/*2.使用IQueryable.Cast<TResult> 方法 ：将IQueryable元素转换为指定的类型*/

List<object> list = new List<object>() { "Apple", "Pear", "Orange" };

            IEnumerable<string> T = list.AsQueryable().Cast<string>();

            foreach (var t in T)
            {
                Console.WriteLine(t);           
            }

/*3.Enumerable.Concat(TSource)方法： 连接两个序列*/

 class pet
        {
            public string name { get; set; }
            public int age { get; set; }

            static pet[] getDogs()
            {
                pet[] dogs ={
                               new pet{name="aa",age=2},
                               new pet{name="bb",age=3},
                               new pet{name="cc",age=4},
                            };
                return dogs;
            }
            static pet[] getCats()
            {
                pet[] cats ={
                                 new pet{name="dd",age=3},
                                 new pet{name="ee",age=5},
                                 new pet{name="ff",age=6},
                             };
                return cats;
            }
            static void Main()
            { 
                pet [] cats=getCats();
                pet [] dogs=getDogs();

                IEnumerable<string> query = cats.Select(cat => cat.name).Concat(dogs.Select(dog => dog.name));
                //下面的这个方法也可以
                //IEnumerable<string> query = cats.AsQueryable().Select(cat => cat.name).Concat(dogs.Select(dog => dog.name));
                foreach (var v in query)
                {
                    Console.WriteLine(v);
                }                            
            }
        }

/*4.Enumerable.Distinct(TSource) 方法：返回非重复的元素*/

            List<int> list = new List<int> { 12, 34, 56, 34, 89, 9, 12 };
            IEnumerable<int> query = list.Distinct();
            foreach (var v in query)
            {
                Console.WriteLine(v);
            }

/*5.Enumerable.GroupBy<TSource,TKey> 方法：根据指定的键选择器函数对序列中的元素进行分组*/

        public string name { get; set; }
        public int age { get; set; }

        static void Main()
        {
            List<pet> pets = new List<pet> { 
                                             new pet{name="aa",age=24},
                                             new pet{name="bb",age=28},
                                             new pet{name="cc",age=25},
                                             new pet{name="dd",age=28},
                                             new pet{name="ee",age=24},
                                             new pet{name="ff",age=28}
                                            };
            var query = pets.AsQueryable().GroupBy(p => p.age);

            foreach (var v in query)
            {
                Console.WriteLine("key:{0},numbers of pet:{1}", v.Key, v.Count());
            }