这是表格:
([TeamA],[TeamB],[Win],[date])
('KKR','HYD','KKR',1),
('KKR','MUM','MUM',2),
('RCB','HYD','HYD',3),
('DEL','PUB','PUB',4),
('RR','PUB','RR',4),
('RR','DEL','RR',5),
('RCB','CSK','CSK',6),
('RR','CSK','RR',7),
('CSK','MUM','MUM',7),
('MUM','DEL','MUM',8),
('HYD','PUNE','PUNE',9),
('PUB','DEL','DEL',9),
('KKR','DEL','KKR',10),
('KKR','RCB','KKR',10)
所需答案应该是连胜 3 场的球队和计数。例如,RR 和 MUM 连续 3 次获胜。 KKR 有 3 场胜利,但是如果我们看到日期列,它不是连续 3 场,因此 KKR 不应该出现在答案中,输出应该是
RR 1
MUM 1
最佳答案
我的方法(可能可以用更干净的方式完成):
WITH cte AS
(
SELECT TeamA AS team FROM #tab
UNION
SELECT TeamB FROM #tab
), cte2 AS
(
SELECT c.team
,[opponent] = CASE WHEN c.team = t.teamA THEN t.teamB ELSE t.teamA END
,t.[win]
,t.[day]
,[is_winner] = CASE WHEN c.team = t.[win] THEN 1 ELSE 0 END
FROM cte c
JOIN #tab t
ON c.team = t.teamA
OR c.team = t.teamB
), cte3 AS
(
SELECT team, [day], [is_winner],
r = ROW_NUMBER() OVER (PARTITION BY team ORDER BY [day])
FROM cte2
), cte4 AS
(
SELECT team, Length = MAX(r) - MIN(r) + 1
FROM (SELECT team, r
,rn=r-ROW_NUMBER() OVER (PARTITION BY team ORDER BY r)
FROM cte3
WHERE is_winner = 1) a
GROUP BY team, rn
)
SELECT team, SUM(Length/3) AS [Number_of_hat_tricks]
FROM cte4
WHERE Length >= 3
GROUP BY team;
输出:
╔══════╦══════════════════════╗
║ team ║ Number_of_hat_tricks ║
╠══════╬══════════════════════╣
║ MUM ║ 1 ║
║ RR ║ 1 ║
╚══════╩══════════════════════╝
它是如何工作的:
- cte - 获取所有团队
- cte2 - 为每个团队找到对手并检查团队是否获胜
- cte3 - 添加连续数字
- cte4 - 计算每个岛屿的长度
- 最终 - 获得 >= 3 的岛屿并将它们相加(整数除法用于将行中的 6 场胜利计为 2,将行中的 9 场胜利计为 3,...)
最后一个想法:
最后一列中的值对于同一团队必须是唯一的:
('RR','CSK','RR',7)
('CSK','MUM','MUM',7)
CSK vs RR - 7
CSK vs MUM - 7
使用当前数据不可能以稳定的方式对其进行排序。所以它应该是日期和时间部分:
CSK vs RR 2015-12-07 10:00
CSK vs MUM 2015-12-07 21:00 -- now we know that it is the second match
关于sql - 如何取得三连胜,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34309320/