計數最大數量

我有一個球員，結果和ID表：計數最大數量

Player | Result | ID 
--------------- 
An  | W  | 1 
An  | W  | 1 
An  | L  | 0 
An  | W  | 1 
An  | W  | 1 
An  | W  | 1 
Ph  | L  | 0 
Ph  | W  | 1 
Ph  | W  | 1 
Ph  | L  | 0 
Ph  | W  | 1

A 'W' 總是會有1的ID，

我需要創建一個查詢，將計算的連續的「W的最大數目爲每個玩家：

Player | MaxWinStreak 
--------------------- 
An  | 3  
Ph  | 2

我試圖用行無界前述但是我只能得到它來計算Ws中的總的最大數量，以及不是c連續地

Select 
    t2.player 
    ,max(t2.cumulative_wins) As 'Max' 

    From 

    ( Select 
      t.Player 
      ,Sum(ID) Over (Partition By t.Result,t.player 
      Order By t.GameWeek Rows Unbounded Preceding) As cumulative_wins 

     From 
      t 

      ) t2 

    Group By 
    t2.player

是否有不同的方法我可以採取？

來源

2017-09-26 PeterH

你需要一個列指定的順序。 –

是否有任何存儲贏利和虧損順序的列？ – TechDo

@GordonLinoff我很高興添加任何額外的列，不知道如何去這個壽 – PeterH

您需要一列來指定排序。 SQL表格代表無序集合。在下面的查詢中，?表示此列。

您可以使用行數之差來獲得各連勝：

select player, count(*) as numwins 
from (select t.*, 
      row_number() over (partition by player order by ?) as seqnum, 
      row_number() over (partition by player, result order by ?) as seqnum_r 
     from t 
    ) t 
where result = 'W' 
group by player, (seqnum - seqnum_r);

然後，您可以得到最大的：

select player, max(numwins) 
from (select player, count(*) as numwins 
     from (select t.*, 
        row_number() over (partition by player order by ?) as seqnum, 
        row_number() over (partition by player, result order by ?) as seqnum_r 
      from t 
      ) t 
     where result = 'W' 
     group by player, (seqnum - seqnum_r) 
    ) pw 
group by player;

來源

2017-09-26 12:44:49

你也許可以使用'（SELECT 1）'order？ – DhruvJoshi

計數最大數量

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