这是表格：
@Pathvariable所需的答案应该是连续3次获胜的球队和计数。这里例如RR和MUM连续三次获胜。 KKR有3胜，但如果我们看到日期列不是连续3，那么KKR不应该在答案中，输出应该是
@RequestMapping(value="/{projectName}/job", method=RequestMethod.GET)
@ResponseBody
public String test(@Pathvariable("projectName") String projectName) throws Exception{
...
}
答案 0 :(得分：3)
我的方法（可能以更干净的方式完成）：
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;
的 LiveDemo
输出：
╔══════╦══════════════════════╗
║ team ║ Number_of_hat_tricks ║
╠══════╬══════════════════════╣
║ MUM ║ 1 ║
║ RR ║ 1 ║
╚══════╩══════════════════════╝
工作原理：
最后一个想法：
最后一列中的值必须与同一团队唯一：
('RR','CSK','RR',7)
('CSK','MUM','MUM',7)
CSK vs RR - 7
CSK vs MUM - 7
使用当前数据无法以稳定的方式对其进行排序。所以应该是时间日期部分：
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
答案 1 :(得分：1)
隙和 - 岛屿。计算每个岛的长度。最终计数是岛的长度除以3（整数除法，丢弃小数部分）。
我添加了更多行（团队A和B）来说明4胜A，然后7胜B，然后4胜{{1}再次，这导致A和A的最终计数为2和2。
示例数据
B<强>查询
通常你会在一个单独的表中有一个团队列表，在这里我用DECLARE @T TABLE (TeamA varchar(50), TeamB varchar(50), Win varchar(50), dt int);
INSERT INTO @T (TeamA, TeamB, Win, dt) VALUES
('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),
('A','B','A',11),
('A','B','A',12),
('A','B','A',13),
('A','B','A',14),
('A','B','B',15),
('A','B','B',16),
('A','B','B',17),
('A','B','B',18),
('A','B','B',19),
('A','B','B',20),
('A','B','B',21),
('A','B','A',22),
('A','B','A',23),
('A','B','A',24),
('A','B','A',25);
构建它。 CTE_Teams每连续连胜的数量为CTE_Counts。由于团队可能会出现多次连胜（见团队3-wins-in-a-row），因此可以进一步总结。团队可以按A和TeamA列中的任意顺序列出，因此在TeamB内的WHERE中有一个CROSS APPLY来捕获这两种变体。< / p>
因此，对于每个团队，只选择与该团队相关的行。这是由OR完成的。
然后，经典CROSS APPLY通过使用不同的分区对行进行两次编号。行数的差异给出了组（岛和间隙）。
过滤器gaps-and-islands只留下获胜团队的岛屿。
按WHERE CTE_Teams.Team = CA.Win分组给出了岛屿的大小，即连续获胜的数量。
查询在SQL Server 2008中有效。
CTE_Teams.Team<强>结果
WITH
CTE_Teams
AS
(
SELECT T.TeamA AS Team
FROM @T AS T
UNION -- sic! not ALL
SELECT T.TeamB AS Team
FROM @T AS T
)
,CTE_Counts
AS
(
SELECT
CTE_Teams.Team
--,CA.Win
--,rn1 - rn2 AS GroupNumber
--,COUNT(*) AS GroupSize
,COUNT(*) / 3 AS FinalCount
FROM
CTE_Teams
CROSS APPLY
(
SELECT
T.Win
,T.dt
,ROW_NUMBER() OVER (PARTITION BY CTE_Teams.Team
ORDER BY T.dt, T.TeamA, T.TeamB) AS rn1
,ROW_NUMBER() OVER (PARTITION BY CTE_Teams.Team, T.Win
ORDER BY T.dt, T.TeamA, T.TeamB) AS rn2
FROM @T AS T
WHERE
T.TeamA = CTE_Teams.Team
OR T.TeamB = CTE_Teams.Team
) AS CA
WHERE
CTE_Teams.Team = CA.Win
GROUP BY
CTE_Teams.Team
,CA.Win
,rn1 - rn2
HAVING COUNT(*) / 3 > 0
)
SELECT
CTE_Counts.Team
,SUM(CTE_Counts.FinalCount) AS FinalCount
FROM CTE_Counts
GROUP BY CTE_Counts.Team
ORDER BY CTE_Counts.Team;
答案 2 :(得分：1)
在不使用CTE的情况下解决此问题的另一种方法是：
create table #a
(
teama varchar(10), teamb varchar(10), win varchar(10), dat int)
insert into #a
values
('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);
select
team,
win,
row_number() over (partition by team order by dat) matchnum
into #res
from
(
select teamA team, case when teamA = win then 1 else 0 end as win, dat
from #a
union all
select teamB team, case when teamB = win then 1 else 0 end , dat
from #a
)A
order by team,dat
select
match1.team, count(*)/3 + 1 cntHatricks
from #res match1 join #res match2
on match1.team = match2.team and match2.matchnum = match1.matchnum+1
join #res match3 on match1.team = match3.team and match3.matchnum = match1.matchnum+2
where
match1.win = 1 and match2.win = 1 and match3.win = 1
group by match1.team
<强>输出
+------+-------------+
| Team | cntHatricks |
+------+-------------+
| MUM | 1 |
| RR | 1 |
+------+-------------+
答案 3 :(得分：1)
尝试这个简单的查询
select tm,sum(case when win=tm then 1 else 0 end)/3 hattrick from @a a
inner join (select teama tm from @a union select teamb from @a) t on a.win=t.tm
group by tm
having count(distinct dat)>2
数据
declare @a table
(teama varchar(10), teamb varchar(10), win varchar(10), dat int)
insert into @a
values
('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)
答案 4 :(得分：0)
用abc作为（选择，案例茶时获胜然后茶叶其他茶叶结束从ipl输掉）
（选择b.win，计数（hatrick）/ 3作为数字
来自ipl b加入
（选择a.win，计数（）为hatrick
来自ipl a加入ipl b on a.win = b.win
其中a.day＆lt; b.day和a.day＆gt;所有
（选择abc.day
从abc加入ipl a on a.win = abc.lose
在哪里abc.lose = b.win和abc.day＆lt; b.day）
a.win小组）d
在b.win = d.win
其中hatrick＆gt; = 3组by b.win）
嘿伙计们，我使用过这个版本。如果这个
中有任何故障，请告诉我