我在sql server中有一个表，它包含像
这样的数据userID amount startdate enddate game Result
-------------------------------------------------------------------------------
68838 51.00 2014-05-29 15:41:41.167 2014-05-29 15:41:41.167 1 W
68838 51.00 2014-05-29 15:42:30.757 2014-05-29 15:42:30.757 1 W
68838 -0.50 2014-05-31 16:57:31.033 2014-05-31 16:57:31.033 1 L
68838 -0.50 2014-05-31 17:05:31.023 2014-05-31 17:05:31.023 1 L
68838 -0.50 2014-05-31 17:22:03.857 2014-05-31 17:22:03.857 1 L
68838 0.42 2014-05-31 17:26:32.570 2014-05-31 17:26:32.570 1 W
68838 0.42 2014-05-31 17:34:45.330 2014-05-31 17:34:45.330 1 W
68838 0.42 2014-05-31 17:38:44.107 2014-05-31 17:38:44.107 1 W
68838 0.42 2014-05-31 17:42:12.790 2014-05-31 17:42:12.790 1 W
434278 0.42 2014-05-31 16:57:31.033 2014-05-31 16:57:31.033 1 W
434278 0.42 2014-05-31 17:05:31.023 2014-05-31 17:05:31.023 1 W
434278 0.42 2014-05-31 17:22:03.857 2014-05-31 17:22:03.857 1 W
434278 -0.50 2014-05-31 17:26:32.570 2014-05-31 17:26:32.570 1 L
434278 -0.50 2014-05-31 17:34:45.330 2014-05-31 17:34:45.330 1 L
434278 -0.50 2014-05-31 17:38:44.107 2014-05-31 17:38:44.107 1 L
434278 -0.50 2014-05-31 17:42:12.790 2014-05-31 17:42:12.790 1 L
434278 0.42 2014-05-31 17:46:40.723 2014-05-31 17:46:40.723 1 W
434278 -0.50 2014-05-31 17:51:26.190 2014-05-31 17:51:26.190 1 L
434278 0.42 2014-05-31 17:55:32.870 2014-05-31 17:55:32.870 1 W
434278 -4.00 2014-05-31 18:06:54.937 2014-05-31 18:06:54.937 1 L
434278 -2.00 2014-05-31 18:19:29.483 2014-05-31 18:19:29.483 1 L
我希望结果看起来像这样，显示每个用户的最长连胜
UserId StartDate Enddate Streak amount
--------------------------------------------------------------------
68838 2014-05-31 17:26:32:570 2014-05-31 17:42:12:570 4 1.68
434278 2014-05-31 16:57:31:033 2014-05-31 17:22:03:857 3 1.26
答案 0 :(得分：2)
免责声明：格伦的答案是一个很好的答案，并为你做了很多繁重的工作，但它并没有完全按照你的要求去做。我打算发一个我一直在努力的答案，但是当我来添加它时，看到Glenn做重物的方式比我做的方式更好，所以我重新设计了我的答案，包括他的方式正在做。我会敦促你接受他的答案，而不是我的答案。
以下内容应完全符合您的要求。
SELECT
Userid,
Min_StartDate as StartDate,
Max_EndDate as EndDate,
max_group_count as Streak,
sum_Amount as Amount
FROM (
SELECT
*,
-- we want the earliest maximum streak
max(Min_StartDate) OVER (PARTITION BY userid) as Earliest_StartDate
FROM (
SELECT
*,
-- we want the maximum streak
max(max_group_count) OVER (PARTITION BY userid) as MAX_Streak
FROM (
SELECT DISTINCT
userid,
-- Calculate this streak
COUNT(grouping) OVER (PARTITION BY userid, grouping
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING
AND UNBOUNDED FOLLOWING ) as max_group_count
-- Calcualte the start date of this streak
,MIN(StartDate) OVER (PARTITION BY userid, grouping
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING
AND UNBOUNDED FOLLOWING ) as Min_StartDate
-- Calcualte the end date of this streak
,MAX(EndDate) OVER (PARTITION BY userid, grouping
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING
AND UNBOUNDED FOLLOWING ) as Max_EndDate
-- Calcualte the total amount
,SUM(Amount) OVER (PARTITION BY userid, grouping
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING
AND UNBOUNDED FOLLOWING ) as Sum_Amount
FROM ( SELECT *
-- Assign a group number to the streak, so we can group by it
,SUM(CASE WHEN result <> prev_result THEN 1 ELSE 0 END) OVER
(PARTITION BY userid ORDER BY startdate) AS grouping
FROM ( SELECT *
-- We want to look at the previous record to determin when the
-- winning/loosing streak starts and ends
,COALESCE(LAG(result) OVER
(PARTITION BY userid ORDER BY startdate), result) AS prev_result
FROM game
) a
WHERE result = 'W'
) b
) c
) d
WHERE
Max_Group_Count = Max_Streak
) e
WHERE
Min_StartDate = Earliest_StartDate
这个输出是：
| USERID | STARTDATE | ENDDATE | STREAK | AMOUNT |
|--------|----------------------------|----------------------------|--------|--------|
| 68838 | May, 31 2014 17:26:32+0000 | May, 31 2014 17:42:12+0000 | 4 | 1.68 |
| 434278 | May, 31 2014 16:57:31+0000 | May, 31 2014 17:22:03+0000 | 3 | 1.26 |
如果您喜欢，我已将此作为您可以玩的sql小提琴：http://sqlfiddle.com/#!6/32777/36/0
答案 1 :(得分：0)
以下是一些让您入门的建议。您可以使用内部查询并开始解决问题。基本上，它首先在显示先前结果的每一行中添加一个额外的列。然后，每次前一个结果与当前结果不同时，请将其视为组切换。每个分组都有自己的编号（基于0，并且在用户标识的上下文中）。丢掉＆＃39; L＆＃39;分组。现在，您对每个用户的最大计数分组感兴趣。
SELECT *
,COUNT(grouping) OVER (PARTITION BY userid, grouping
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING
AND UNBOUNDED FOLLOWING ) as max_group_count
FROM ( SELECT *
,SUM(CASE WHEN result <> prev_result THEN 1 ELSE 0 END) OVER
(PARTITION BY userid ORDER BY startdate) AS grouping
FROM ( SELECT *
,COALESCE(LAG(result) OVER
(PARTITION BY userid ORDER BY startdate), result) AS prev_result
FROM game
) x
WHERE result = 'W'
) y
结果：
userid | startdate | enddate | result | prev_result | grouping | max_group_count
--------+-------------------------+-------------------------+--------+-------------+----------+-----------------
68838 | 2014-05-29 15:41:41.167 | 2014-05-29 15:41:41.167 | W | W | 0 | 2
68838 | 2014-05-29 15:42:30.757 | 2014-05-29 15:42:30.757 | W | W | 0 | 2
68838 | 2014-05-31 17:26:32.57 | 2014-05-31 17:26:32.57 | W | L | 1 | 4
68838 | 2014-05-31 17:34:45.33 | 2014-05-31 17:34:45.33 | W | W | 1 | 4
68838 | 2014-05-31 17:38:44.107 | 2014-05-31 17:38:44.107 | W | W | 1 | 4
68838 | 2014-05-31 17:42:12.79 | 2014-05-31 17:42:12.79 | W | W | 1 | 4
434278 | 2014-05-31 16:57:31.033 | 2014-05-31 16:57:31.033 | W | W | 0 | 3
434278 | 2014-05-31 17:05:31.023 | 2014-05-31 17:05:31.023 | W | W | 0 | 3
434278 | 2014-05-31 17:22:03.857 | 2014-05-31 17:22:03.857 | W | W | 0 | 3
434278 | 2014-05-31 17:46:40.723 | 2014-05-31 17:46:40.723 | W | L | 1 | 1
434278 | 2014-05-31 17:55:32.87 | 2014-05-31 17:55:32.87 | W | L | 2 | 1
(11 rows)
答案 2 :(得分：0)
在SQLServer 2008中，SUM无法使用OVER(ORDER BY ...)，这会使查询变得更复杂，但并非不可能
;WITH myID AS (
SELECT userID, amount, startdate, enddate, game, Result
, ID = Row_Number() OVER (Partition By userID ORDER BY startdate)
FROM Table1 t1
), SR AS (
SELECT t1.userID, t1.startdate, t1.enddate, t1.Result, t1.amount
, SUM(CASE WHEN t1.Result <> COALESCE(t2.Result, t1.Result)
THEN 1
ELSE 0 END) SC
FROM myID t1
LEFT JOIN myID t2 ON t1.userID = t2.userID AND t1.ID >= t2.ID
GROUP BY t1.userID, t1.startdate, t1.enddate, t1.Result, t1.amount
), SL AS (
SELECT userID, Result, SC, Count(1) Streak
, Row_Number() OVER (PARTITION BY userID ORDER BY Count(1) DESC) Pos
FROM SR
WHERE Result = 'W'
GROUP BY userID, Result, SC
)
SELECT p.userID
, MIN(p.startdate) startdate
, MAX(p.enddate) enddate
, l.Streak
, SUM(p.Amount) Amount
FROM SR p
INNER JOIN SL l ON p.userID = l.userID AND p.SC = l.SC
WHERE l.Pos = 1
GROUP BY p.userID, l.Streak
myID CTE生成数据的整数ID，以简化下一个JOIN的{{1}}条件，如果问题中显示的数据不是整个表格，并且已经存在一个具有相同效果的列，应删除此CTE
CTE（StreakRank）SR将数据添加到Streak计数器CTE，使用三角形SC为每条条纹生成排名，它不是密集排名，它是只是用来分组的东西。
JOIN（StreakLength）SL获得每个连胜的连胜长度，并按长度创建排名。
主查询将所有内容放在一起：从CTE usedID和SL获取JOIN的最长条纹以获取详细信息。