age*_*ega 27 sql t-sql select sql-server-2008
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
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如何选择已加星标的行,Value即已更改的位置?基本上我试图找到时间Value已经改变,所以我可以根据这些时间间隔做其他查询.解决方案不应该依赖于知道Value或Time提前.
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.
我尝试在这里调整解决方案http://blog.sqlauthority.com/2011/11/24/sql-server-solution-to-puzzle-simulate-lead-and-lag-without-using-sql-server-2012-解析函数/但我的查询在一小时后没有完成!我认为连接将行大小爆炸到无法管理的东西(或者我搞砸了).
我可以使用C#代码和多个db调用来解决这个问题,但它似乎可以在表值函数或SP中完成,这样可以更好.
此外,Value如果更容易,只有增加的解决方案才行.
Aar*_*and 31
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;
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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,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;
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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
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小智 12
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
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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
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