我有这张桌子：
table session(
ID number,
SessionID VarChar,
Date,
Filter
)
此表包含搜索信息，如下所示：
ID SessionID Date filter
4 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 meagPixel=5
6 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 brand=Canon
7 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 brand=Canon&meagPixel=12.1
8 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 brand=Canon
10 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 brand=Nikon
12 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 meagPixel=12.1
13 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 meagPixel=12.1&opticalZoom=True
14 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 meagPixel=12.1&opticalZoom=True&brand=Panasonic
16 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 price=500.00
18 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 price=499.00
19 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 price=499.00&brand=Olympus
21 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 zoomRange=2000
22 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 zoomRange=2000&brand=Leica
23 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 zoomRange=2000&brand=Leica&price=1995.00
24 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True
25 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True&meagPixel=16.2
26 peqq421gaspts3nuulq5mwcq 24/05/2012 13:50 zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True&meagPixel=16.2&weight=345
27 peqq421gaspts3nuulq5mwcq 24/05/2012 13:58 zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True&meagPixel=16.2
41 poiq41111spts00000q5aaaa 27/05/2012 13:48 meagPixel=5
我想获得独特的搜索。唯一的搜索是：
由于ASP.NET不保证SessionID是唯一的（SessionID，Date）是唯一的。
我没有太远：
SELECT MAX(Filter)
FROM Session
GROUP BY SessionID
BTW我给出的示例表数据的结果应该返回：
ID SessionID Date filter
4 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 meagPixel=5
7 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 brand=Canon&meagPixel=12.1
10 peqq421gaspts3nuulq5mwcq 24/05/2012 13:48 brand=Nikon
14 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 meagPixel=12.1&opticalZoom=True&brand=Panasonic
16 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 price=500.00
19 peqq421gaspts3nuulq5mwcq 24/05/2012 13:49 price=499.00&brand=Olympus
26 peqq421gaspts3nuulq5mwcq 24/05/2012 13:50 zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True&meagPixel=16.2&weight=345
41 poiq41111spts00000q5aaaa 27/05/2012 13:48 meagPixel=5
感谢您的帮助和指导。
答案 0 :(得分：1)
@GarethD - 架构和插入查询的Tx。 我尝试过略有不同的方法。我不确定这是否适用于所有情况。它适用于mysql和mssql。
select *
from tsession t1
where not exists (
select *
from tsession t2
where t2.filter like concat(t1.filter,'%')
and t1.filter<>t2.filter
and t1.sessionid=t2.sessionid)
order by id;
根据问题的要求提供准确的结果。
答案 1 :(得分：0)
要获得最长的搜索过滤器，您需要执行以下操作：
select s.*
from (select s.*,
row_number() over (partition by sessionid order by len desc) as rownum
from (select s.*, len(filter) as len
from session s
) s
) s
where rownum = 1
我正在使用Windows函数执行此操作。您可以使用聚合和连接来执行相同的操作。
但是，您说会话不是真正的标识符。会话/过滤器是。以下查询几乎可以获得您想要的内容：
select s.*
from (select s.*,
row_number() overo over (partition by sessionid, filter
order by len desc) as rownum
from (select s.*, len(filter) as len
from session s
) s
) s
where rownum = 1
（唯一的变化是分区子句包含过滤器。）
您可能有重复项。如果你想要所有重复项，那么稍微不同的查询就可以了。
答案 2 :(得分：0)
首先，您的示例数据中存在错误我认为第25,26和27行应该都出现在您的最终数据中。 27肯定应该是因为它是会话ID和日期组合的唯一条目。
假设上述情况正确，那么我认为我已经正确地建立了你的逻辑。
步骤1是为每个过滤器定义第一个搜索词，以及在会话中出现的顺序：
;WITH CTE AS
( SELECT *,
SUBSTRING(Filter, 1, CASE WHEN CHARINDEX('&', Filter) = 0 THEN LEN(Filter) ELSE CHARINDEX('&', Filter) - 1 END) [FirstTerm],
FROM Session
)
下一步是确定每次搜索是新搜索还是前一次搜索的延续。这是通过在会话中获取Previous搜索词（为什么SessionOrder在最后一个CTE中定义）并确定第一个搜索词是否相同来完成的。
, CTE2 AS
( SELECT T1.*,
CASE WHEN T1.SessionOrder = 1 OR T2.SessionOrder IS NOT NULL THEN 1 ELSE 0 END [NewSearch]
FROM CTE T1
LEFT JOIN CTE T2
ON T1.SessionID = T2.SessionID
AND T1.Date = T2.Date
AND T1.FirstTerm != T2.FirstTerm
AND T1.SessionOrder = T2.SessionOrder + 1
)
接下来，每个新搜索都需要在会话中拥有自己的排名，用于分组purpuses。然后，您定义了规则（SessionID，Date和First Search术语的唯一组合），然后您可以根据过滤器的长度对唯一组合中的每个项目进行排序：
, CTE3 AS
( SELECT *,
ROW_NUMBER() OVER(PARTITION BY SessionID, Date, ISNULL(SearchNumber, 0) ORDER BY LEN(Filter) DESC) [SearchOrder]
FROM CTE2 T1
OUTER APPLY
( SELECT SUM(NewSearch) [SearchNumber]
FROM CTE2 T2
WHERE T1.SessionOrder >= T2.SessionOrder
AND T1.SessionID = T2.SessionID
AND T1.Date = T2.Date
) c
)
最后，您需要做的就是将结果限制为SessionID，日期和第一个过滤条件的每个组合的最长搜索词：
SELECT ID, SessionID, Date, Filter
FROM CTE3
WHERE SearchOrder = 1
ORDER BY ID
通常情况下，我会把这些全部放在SQLFiddle上，而不是在这里发布一个完整的工作示例，但它今天似乎没有工作。所以这是我用来测试你的数据的完整SQL：
CREATE TABLE #Session (ID INT, SessionID VARCHAR(50), Date DATETIME, Filter VARCHAR(200))
INSERT INTO #Session VALUES
(2, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:48', 'brand=Canon'),
(4, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:48', 'meagPixel=5'),
(6, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:48', 'brand=Canon'),
(7, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:48', 'brand=Canon&meagPixel=12.1'),
(8, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:48', 'brand=Canon'),
(10, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:48', 'brand=Nikon'),
(12, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:48', 'meagPixel=12.1'),
(13, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:48', 'meagPixel=12.1&opticalZoom=True'),
(14, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'meagPixel=12.1&opticalZoom=True&brand=Panasonic'),
(16, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'price=500.00'),
(18, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'price=499.00'),
(19, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'price=499.00&brand=Olympus'),
(21, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'zoomRange=2000'),
(22, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'zoomRange=2000&brand=Leica'),
(23, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'zoomRange=2000&brand=Leica&price=1995.00'),
(24, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True'),
(25, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:49', 'zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True&meagPixel=16.2'),
(26, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:50', 'zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True&meagPixel=16.2&weight=345'),
(27, 'peqq421gaspts3nuulq5mwcq', '24/05/2012 13:58', 'zoomRange=2000&brand=Leica&price=1995.00&opticalZoom=True&meagPixel=16.2'),
(41, 'poiq41111spts00000q5aaaa', '27/05/2012 13:48', 'meagPixel=5')
;WITH CTE AS
( SELECT *,
SUBSTRING(Filter, 1, CASE WHEN CHARINDEX('&', Filter) = 0 THEN LEN(Filter) ELSE CHARINDEX('&', Filter) - 1 END) [FirstTerm],
FROM #Session
), CTE2 AS
( SELECT T1.*,
CASE WHEN T1.SessionOrder = 1 OR T2.SessionOrder IS NOT NULL THEN 1 ELSE 0 END [NewSearch]
FROM CTE T1
LEFT JOIN CTE T2
ON T1.SessionID = T2.SessionID
AND T1.Date = T2.Date
AND T1.FirstTerm != T2.FirstTerm
AND T1.SessionOrder = T2.SessionOrder + 1
), CTE3 AS
( SELECT *,
ROW_NUMBER() OVER(PARTITION BY SessionID, Date, ISNULL(SearchNumber, 0) ORDER BY LEN(Filter) DESC) [SearchOrder]
FROM CTE2 T1
OUTER APPLY
( SELECT SUM(NewSearch) [SearchNumber]
FROM CTE2 T2
WHERE T1.SessionOrder >= T2.SessionOrder
AND T1.SessionID = T2.SessionID
AND T1.Date = T2.Date
) c
)
SELECT ID, SessionID, Date, Filter
FROM CTE3
WHERE SearchOrder = 1
ORDER BY ID
DROP TABLE #Session
<强>附录
好的，根据您的结果集，您实际上并不想按日期列进行分组，您只需按第一个搜索词和sessionID分组的长度顺序排列行。
此查询产生与样本数据相同的结果。我在2008 R1中测试了这个，但是没有理由认为它在SQL-Server CE中不起作用。
;WITH CTE AS
( SELECT *,
ROW_NUMBER() OVER(PARTITION BY SessionID, SUBSTRING(Filter, 1, CASE WHEN CHARINDEX('&', Filter) = 0 THEN LEN(Filter) ELSE CHARINDEX('&', Filter) - 1 END) ORDER BY LEN(Filter) DESC) [RowNumber]
FROM Session
)
SELECT *
FROM CTE
WHERE RowNumber = 1
ORDER BY ID
SQL Fiddle最终解决方案