- 向下滚动以查看我添加的编辑 -
所以这是我的情景。我有一个表，每当有人对某些数据进行更改时都会有一个条目。原因是我们需要能够审核所有更改。
但是，我只想检索用户进行的一系列编辑的最新记录。
所以假设有三个用户，用户A，B和C.
用户A进行了10次更改（表中有10个条目）。 用户B进行了5次更改 用户A进行了3次更改 用户C进行了2次更改
我想要回来的是： C创建的2条记录中的最新记录 A创建的3条记录中的最新记录 B创建的5条记录中的最新记录 A创建的10条记录中的最新记录
总共4行我回来了
这是我尝试过的，但问题是，当LastUpdatedBy发生变化时，RowNum不会回到1：
WITH cte AS
(
SELECT
[LastUpdatedOn]
,[LastUpdatedBy]
,ROW_NUMBER() OVER(PARTITION BY [LastUpdatedBy] ORDER BY [LastUpdatedOn] DESC) [RowNum]
FROM [HistoricalTable]
)
SELECT
[LastUpdatedOn]
,[LastUpdatedBy]
,RowNum
FROM cte
--WHERE RowNum = 1
ORDER BY [LastUpdatedOn] DESC;
这是我得到的输出（**星号代表我想要的行）
LastUpdatedOn LastUpdatedBy RowNum
**2011-06-07 13:07:26.917 629 1**
2011-06-07 12:57:53.700 629 2
2011-06-07 12:57:44.387 629 3
2011-06-07 12:57:34.913 629 4
2011-06-07 12:57:25.040 629 5
2011-06-07 12:57:19.927 629 6
2011-06-07 12:55:17.460 629 7
2011-06-07 12:55:12.287 629 8
2011-06-07 12:30:34.377 629 9
**2011-06-07 11:54:05.727 4 1**
**2011-06-07 11:50:02.723 629 10** (If this number went back to 1, my query would have worked fine)
2011-06-07 11:26:43.053 629 11
2011-06-07 10:54:32.867 629 12
2011-06-07 10:46:32.107 629 13
2011-06-07 10:40:52.937 629 14
**2011-06-07 10:39:50.880 3 1**
-------------------编辑--------------------
所以我提出了一个解决方案，但它并不是很优雅，也不确定我是否喜欢它，但它可以解决问题。这可能会让您更好地理解我想要完成的任务。
DECLARE @temp AS TABLE(LastUpdatedOn datetime, LastUpdatedBy int null, RowNum int);
DECLARE @newTable AS TABLE(LastUpdatedOn datetime, LastUpdatedBy int null);
DECLARE @lastUserId int = 0;
INSERT INTO @temp
SELECT
[LastUpdatedOn]
,[LastUpdatedBy]
,ROW_NUMBER() OVER(ORDER BY [LastUpdatedOn] DESC) [RowNum]
FROM [HistoricalTable]
DECLARE @totalRecords int;
SELECT @totalRecords = COUNT(*) FROM @temp;
DECLARE @counter int = 0;
WHILE @counter <= @totalRecords BEGIN
SET @counter = @counter + 1;
INSERT INTO @newTable
SELECT LastUpdatedOn, LastUpdatedBy
FROM @temp
WHERE RowNum = @counter AND (@lastUserId != LastUpdatedBy OR (LastUpdatedBy IS NULL));
SELECT @lastUserId = LastUpdatedBy FROM @temp WHERE RowNum = @counter;
END
SELECT * FROM @newTable;
返回的数据：
LastUpdatedOn LastUpdatedBy
2011-06-07 13:07:26.917 629
2011-06-07 11:54:05.727 4
2011-06-07 11:50:02.723 629
2011-06-07 10:39:50.880 3
答案 0 :(得分：5)
;with cte as
(
select *,
row_number() over(order by LastUpdatedOn) as rn
from HistoricalTable
)
select C1.LastUpdatedOn,
C1.LastUpdatedBy
from cte as C1
left outer join cte as C2
on C1.rn = C2.rn-1
where C1.LastUpdatedBy <> coalesce(C2.LastUpdatedBy, 0)
按LastUpdatedOn为每个行顺序创建行号，并加入下一行并比较LastUpdatedBy是否已更改。
谨防这个coalesce(C2.LastUpdatedBy, 0)。它是获取最后一行，0需要是一个不用作LastUpdatedBy的整数值。
答案 1 :(得分：2)
不确定我是否遗漏了你问题中的内容，但是下面的SQL没有回答这个问题？
declare @HistoricalTable table (LastUpdatedOn datetime, LastUpdatedBy int);
insert into @HistoricalTable (LastUpdatedOn, LastUpdatedBy) values
('2011-06-07 13:07:26.917', 629),('2011-06-07 12:57:53.700', 629),
('2011-06-07 12:57:44.387', 629),('2011-06-07 12:57:34.913', 629),
('2011-06-07 12:57:25.040', 629),('2011-06-07 12:57:19.927', 629),
('2011-06-07 12:55:17.460', 629),('2011-06-07 12:55:12.287', 629),
('2011-06-07 12:30:34.377', 629),('2011-06-07 11:54:05.727', 4),
('2011-06-07 11:50:02.723', 629),('2011-06-07 11:26:43.053', 629),
('2011-06-07 10:54:32.867', 629),('2011-06-07 10:46:32.107', 629),
('2011-06-07 10:40:52.937', 629),('2011-06-07 10:39:50.880', 3);
select
latest.*
from
(
select *, rank() over (partition by LastUpdatedBy order by LastUpdatedOn desc) as UpdateRank
from @HistoricalTable
) latest
where
latest.UpdateRank = 1
order by
latest.LastUpdatedBy;
LastUpdatedOn LastUpdatedBy UpdateRank
2011-06-07 10:39:50.880 3 1
2011-06-07 11:54:05.727 4 1
2011-06-07 13:07:26.917 629 1
答案 2 :(得分：1)
今天早上让我感到震惊的是这是一个岛屿问题。这是我的解决方案：
CREATE TABLE #tmp (
LastUpdatedBy INT,
LastUpdatedOn DATETIME
)
INSERT INTO #tmp
( LastUpdatedOn, LastUpdatedBy )
VALUES ( '2011-06-07 13:07:26.917', 629 ),
( '2011-06-07 12:57:53.700', 629 ),
( '2011-06-07 12:57:44.387', 629 ),
( '2011-06-07 12:57:34.913', 629 ),
( '2011-06-07 12:57:25.040', 629 ),
( '2011-06-07 12:57:19.927', 629 ),
( '2011-06-07 12:55:17.460', 629 ),
( '2011-06-07 12:55:12.287', 629 ),
( '2011-06-07 12:30:34.377', 629 ),
( '2011-06-07 11:54:05.727', 4 ),
( '2011-06-07 11:50:02.723', 629 ),
( '2011-06-07 11:26:43.053', 629 ),
( '2011-06-07 10:54:32.867', 629 ),
( '2011-06-07 10:46:32.107', 629 ),
( '2011-06-07 10:40:52.937', 629 ),
( '2011-06-07 10:39:50.880', 3 ) ;
WITH cte
AS ( SELECT [LastUpdatedOn],
[LastUpdatedBy],
ROW_NUMBER() OVER ( PARTITION BY [LastUpdatedBy] ORDER BY [LastUpdatedOn] DESC ) - ROW_NUMBER() OVER ( ORDER BY [LastUpdatedOn] DESC ) AS [Island]
FROM #tmp
),
cte2
AS ( SELECT *,
ROW_NUMBER() OVER ( PARTITION BY [Island] ORDER BY [LastUpdatedOn] DESC ) AS [rn]
FROM cte
)
SELECT [LastUpdatedOn],
[LastUpdatedBy]
FROM cte2
WHERE [rn] = 1
ORDER BY [LastUpdatedOn] DESC ;
这里的“技巧”是要注意，如果你在分区内和整个集合中跟踪row_number，那么当分区改变时，两者之间的差异将会改变。
答案 3 :(得分：0)
这完全未经测试，但它可能构成工作解决方案的基础：
SELECT
[Outer].[LastUpdatedOn],
[Outer].[LastUpdatedBy]
FROM [HistoricalTable] AS [Outer]
WHERE NOT EXISTS
(
SELECT *
FROM [HistoricalTable] AS [Middle]
WHERE [Middle].[LastUpdatedBy] = [Outer].[LastUpdatedBy]
AND [Middle].[LastUpdatedOn] > [Outer].[LastUpdatedOn]
AND [Middle].[LastUpdatedOn] <= ISNULL(
(
SELECT
MIN([Inner].[LastUpdatedOn])
FROM [HistoricalTable] AS [Inner]
WHERE [Inner].[LastUpdatedBy] != [Outer].[LastUpdatedBy]
AND [Inner].[LastUpdatedOn] > [Outer].[LastUpdatedOn]
), [Middle].[LastUpdatedOn])
)
即使这种方法有效，假设你不仅仅有少数几行，性能也可能很糟糕。
对于表中的每一行，它确保同一用户在上下文行和最新行之间不存在任何其他行，这些行比链接到不同用户的上下文行更新。