SQL查询根据列值变化查找最早日期
我有一个问题,我需要从按列分组的表中获取最早的日期值,但按顺序分组.
I have a problem where I need to get the earliest date value from a table grouped by a column, but sequentially grouped.
这是一个示例表:
if object_id('tempdb..#tmp') is NOT null
DROP TABLE #tmp
CREATE TABLE #tmp
(
UserID BIGINT NOT NULL,
JobCodeID BIGINT NOT NULL,
LastEffectiveDate DATETIME NOT NULL
)
INSERT INTO #tmp VALUES ( 1, 5, '1/1/2010')
INSERT INTO #tmp VALUES ( 1, 5, '1/2/2010')
INSERT INTO #tmp VALUES ( 1, 6, '1/3/2010')
INSERT INTO #tmp VALUES ( 1, 5, '1/4/2010')
INSERT INTO #tmp VALUES ( 1, 1, '1/5/2010')
INSERT INTO #tmp VALUES ( 1, 1, '1/6/2010')
SELECT JobCodeID, MIN(LastEffectiveDate)
FROM #tmp
WHERE UserID = 1
GROUP BY JobCodeID
DROP TABLE [#tmp]
此查询将返回 3 行,其中包含最小值.
This query will return 3 rows, with the min value.
1 2010-01-05 00:00:00.000
5 2010-01-01 00:00:00.000
6 2010-01-03 00:00:00.000
我正在寻找的是该组是连续的并返回多个 JobCodeID,如下所示:
What I am looking for is for the group to be sequential and return more than one JobCodeID, like this:
5 2010-01-01 00:00:00.000
6 2010-01-03 00:00:00.000
5 2010-01-04 00:00:00.000
1 2010-01-05 00:00:00.000
这可以不用游标吗?
推荐答案
SELECT JobCodeId, MIN(LastEffectiveDate) AS mindate
FROM (
SELECT *,
prn - rn AS diff
FROM (
SELECT *,
ROW_NUMBER() OVER (PARTITION BY JobCodeID
ORDER BY LastEffectiveDate) AS prn,
ROW_NUMBER() OVER (ORDER BY LastEffectiveDate) AS rn
FROM @tmp
) q
) q2
GROUP BY
JobCodeId, diff
ORDER BY
mindate
连续范围在分区和未分区ROW_NUMBERs
之间具有相同的差异.
Continuous ranges have same difference between partitioned and unpartitioned ROW_NUMBERs
.
您可以在 GROUP BY
中使用此值.
You can use this value in the GROUP BY
.
有关其工作原理的更多详细信息,请参阅我博客中的这篇文章:
See this article in my blog for more detail on how it works:
- 对连续范围进行分组
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