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？基本上我正在尝试找到Value已更改的时间，因此我可以根据这些时间间隔进行其他查询。解决方案不应该依赖于提前了解Value或Time。

在我看来，这应该不是很难（但显然对我来说很难！）。

我目前正在使用SQL Server 2008，但如果新窗口/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，但一小时后我的查询没有完成！我认为连接会将行大小分解为无法管理的（或者我搞砸了）。

我可以使用C＃代码和多个db调用解决这个问题，但似乎可以在表值函数或SP中完成，这样可以更好。

此外，如果更容易，只有增加Value的解决方案才行。

Answer 1

我认为这就是你所追求的：

;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......

修改

啊，等一下，价值观上涨和下跌，而不仅仅是......如果是这种情况你可以尝试这么慢的方法：

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

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，如果它改变了，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值不同的所有行不是1和sub2.rn IS NULL用于第一行，因为连接将从id = 2开始。

选择列值已更改的行

3 个答案: