选择列值已更改的行

Question

假设我有下表：

Value    Time
0        15/06/2012 8:03:43 PM
1        15/06/2012 8:03:43 PM     *
1        15/06/2012 8:03:48 PM 
1        15/06/2012 8:03:53 PM
1        15/06/2012 8:03:58 PM     
2        15/06/2012 8:04:03 PM     *
2        15/06/2012 8:04:08 PM
3        15/06/2012 8:04:13 PM     *
3        15/06/2012 8:04:18 PM
3        15/06/2012 8:04:23 PM
2        15/06/2012 8:04:28 PM     *
2        15/06/2012 8:04:33 PM     

如何选择已加星标的行，即Value 已更改的位置？基本上我试图找出Value 发生变化的时间，以便我可以根据这些时间间隔进行其他查询。解决方案不应依赖于提前知道Value 或Time。

在我看来，这应该不是很难（但显然对我来说已经够难了！）。

我目前使用的是 SQL Server 2008，但如果新的窗口/分析功能有用，我可以访问 2012。

我尝试在http://blog.sqlauthority.com/2011/11/24/sql-server-solution-to-puzzle-simulate-lead-and-lag-without-using-sql-server-2012-analytic-function/ 调整解决方案，但一个小时后我的查询没有完成！我认为连接将行大小爆炸到无法管理的程度（或者我搞砸了）。

我可以通过 C# 代码和多个 db 调用来解决这个问题，但似乎可以在表值函数或 SP 中完成，这样会更好。

此外，如果更简单，仅适用于增加 Value 的解决方案也可以。

Answer 1

我们也可以使用子查询来做到这一点

SELECT sub1.value, sub1.time FROM 
  (SELECT *,rn,id FROM 
     (SELECT *,row_number() over (partition by value order by time) AS rn, row_number() over (order by time) AS id FROM x ) order by time) sub1
  LEFT OUTER JOIN 
  (SELECT *,rn,id FROM 
     (SELECT *,row_number() over (partition by value order by time) AS rn, row_number() over (order by time) AS id FROM x ) order by time) sub2
  ON sub1.id = sub2.id + 1 
  WHERE sub1.rn - sub2.rn <> 1 OR sub2.rn IS NULL;

所以，我比较了 2 行的值，如果它发生变化，那么 rn 的差值将不等于 1，否则 rn 值将增加 1，所以，我选择了与下一行的 rn 值差为的所有行not 1 and sub2.rn IS NULL 用于第一行，因为连接将从 id = 2 开始。

Answer 2

DECLARE @x TABLE(value INT, [time] DATETIME)

INSERT @x VALUES
(0,'20120615 8:03:43 PM'),--
(1,'20120615 8:03:43 PM'),--*
(1,'20120615 8:03:48 PM'),--
(1,'20120615 8:03:53 PM'),--
(1,'20120615 8:03:58 PM'),--
(2,'20120615 8:04:03 PM'),--*
(2,'20120615 8:04:08 PM'),--
(3,'20120615 8:04:13 PM'),--*
(3,'20120615 8:04:18 PM'),--
(3,'20120615 8:04:23 PM'),--
(2,'20120615 8:04:28 PM'),--*
(2,'20120615 8:04:33 PM');


; with temp as
(
SELECT 
    value, [time],  lag(value,1,-1) over (order by [time] ) as lastValue
FROM    @x
) 
SELECT 
    [value],[time] 
FROM 
    temp 
WHERE value <> lastValue

结果：

value   time
---------------------------
0   2012-06-15 20:03:43.000
1   2012-06-15 20:03:43.000
2   2012-06-15 20:04:03.000
3   2012-06-15 20:04:13.000
2   2012-06-15 20:04:28.000

Answer 3

我想这就是你所追求的：

;WITH x AS
(
  SELECT value, time, rn = ROW_NUMBER() OVER 
  (PARTITION BY Value ORDER BY Time)
  FROM dbo.table
)
SELECT * FROM x WHERE rn = 1;

如果结果集很大并且没有良好的支持索引，这可能会很慢...

编辑

啊，等一下，值会上升和下降，而不仅仅是上升...如果是这种情况，您可以尝试这种慢得多的方法：

DECLARE @x TABLE(value INT, [time] DATETIME)

INSERT @x VALUES
(0,'20120615 8:03:43 PM'),--
(1,'20120615 8:03:43 PM'),--*
(1,'20120615 8:03:48 PM'),--
(1,'20120615 8:03:53 PM'),--
(1,'20120615 8:03:58 PM'),--
(2,'20120615 8:04:03 PM'),--*
(2,'20120615 8:04:08 PM'),--
(3,'20120615 8:04:13 PM'),--*
(3,'20120615 8:04:18 PM'),--
(3,'20120615 8:04:23 PM'),--
(2,'20120615 8:04:28 PM'),--*
(2,'20120615 8:04:33 PM');

;WITH x AS
(
  SELECT *, rn = ROW_NUMBER() OVER (ORDER BY time)
  FROM @x
)
SELECT x.value, x.[time]
FROM x LEFT OUTER JOIN x AS y
ON x.rn = y.rn + 1
AND x.value <> y.value
WHERE y.value IS NOT NULL;

结果：

value  time
-----  -----------------------
1      2012-06-15 20:03:43.000
2      2012-06-15 20:04:03.000
3      2012-06-15 20:04:13.000
2      2012-06-15 20:04:28.000