C# Linq

无聊一下，感觉有点用记录一下，^_^  用过linq的请绕道...

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List<TestCase> networks = list.GroupBy(o => o.NetWork)                     .Select(group => new TestCase { NetWork = group.Key })                     .ToList();

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List<Test> list = new List<Test>(); list.Add(new Test { score = 10, name = "张君宝" }); list.Add(new Test { score = 20, name = "刘惜君" }); list.Add(new Test { score = 20, name = "刘惜君" }); list.Add(new Test { score = 30, name = "八戒" });   int total = (from temp in list                  where temp.name == "刘惜君"                  select temp.age).Count();   int score = (from temp in list              where temp.name == "刘惜君"             select temp.score).Sum(); this.textBox1.Text = "刘惜君：" + total + "人" + score + "score";   MessageBox.Show(this.textBox1.Text);   <br>//return 刘惜君：2人40score

语法如下：　

from 临时变量 in 实现IEnumerable<T>接口的对象

         where条件表达式

         [orderby 条件]

         [group by 条件]

         select 临时变量中被查询的值

以下内容来源于网络：

  //1,使用LINQ做查询( 表达式写法)
            var res = from m in masterList
                      //from后面设置查询的集合
                      where m.Level > 8 && m.Menpai == "丐帮"
                      //where后面跟上查询的条件
                      select m;//表示m的结果结合返回
            var res = from m in masterList

                      from k in kongfuList
                      where m.Kongfu == k.Name && k.Power > 90
                      select m;

 Join on集合联合查询

            var res = from m in masterList
                      // join...in... 表示要连接的表，on后面为连接条件，等于要用equals，不能用==
                      join k in kongfuList on m.Kongfu equals k.Name
                      where k.Power > 90
                      select new { master = m, kongfu = k };


分组查询 into groups 

把武林高手按照所学功夫分类，看一下那个功夫修炼的人数最多

            var res = from k in kongfuList
                      join m in masterList on k.Name equals m.Kongfu
                      into groups   //分组
                      orderby groups.Count()  // 这个可以获得数量
                      select new { kongfu = k, count = groups.Count() };

来自：https://www.cnblogs.com/lmx282110xxx/p/10798682.html