在linq中，关于SelectMany和Aggregate的用法

1.SelectMany 用于建立 两个集合的对应关系

2.Aggregate 进行迭代（如果不提供种子，那么第一个元素就是种子）

var aa = new string[] { "aa1","aa2" };
var bb = new string[] { "bb1","bb2","bb3"};

//a => bb 表明集合aa 和 结合bb 建立迪科尔乘积
var s1 = aa.SelectMany(a => bb,(a,b) => a + "->"+b)
    //如果提供了种子，种子就是第一个元素
    .Aggregate("Header;",(curr,next) => curr + next + ";")
    .ToArray();
//Header;aa1->bb1;aa1->bb2;aa1->bb3;aa2->bb1;aa2->bb2;aa2->bb3;
Console.WriteLine(s1);

s1 = aa.SelectMany(a => bb, (a, b) => a + "->" + b)
    .Aggregate( (curr, next) => curr + next + ";")
    .ToArray();
//如果没有种子，第一个元素就是种子
//aa1->bb1aa1->bb2;aa1->bb3;aa2->bb1;aa2->bb2;aa2->bb3;
Console.WriteLine(s1);

s1 = aa.SelectMany(a => bb, (a, b) => a + "->" + b)
    .Aggregate((curr, next) => curr + ";" + next )
    .ToArray();
//如果没有种子，第一个元素就是种子
//aa1->bb1;aa1->bb2;aa1->bb3;aa2->bb1;aa2->bb2;aa2->bb3
Console.WriteLine(s1);


string[] fruits = { "apple", "mango", "orange", "passionfruit", "grape" };

// Determine whether any string in the array is longer than "banana".
string longestName =
    fruits.Aggregate("banana",
                    (longest, next) =>next.Length > longest.Length ? next : longest,
                    // Return the final result as an upper case string.
                    fruit => fruit.ToUpper());

Console.WriteLine("The fruit with the longest name is {0}.",longestName);
// The fruit with the longest name is PASSIONFRUIT.

// Determine whether any string in the array is longer than "banana".
//未提供种子，第一个元素就是种子了
longestName = fruits.Aggregate((longest, next) => next.Length > longest.Length ? next : longest);

Console.WriteLine("The fruit with the longest name is {0}.", longestName);
// The fruit with the longest name is PASSIONFRUIT.