hel*_*llo 3 sql-server row-number lag ranking-functions partition-by
我遇到了与此问题非常相似的问题 T-sql Reset Row number on Field Change
这个问题的解决方案很完美,效果很好。除非我尝试使用多个其他“custno”,否则它会崩溃。
我的意思是:
custno moddate who
--------------------------------------------------
581827 2012-11-08 08:38:00.000 EMSZC14
581827 2012-11-08 08:41:10.000 EMSZC14
581827 2012-11-08 08:53:46.000 EMSZC14
581827 2012-11-08 08:57:04.000 EMSZC14
581827 2012-11-08 08:58:35.000 EMSZC14
581827 2012-11-08 08:59:13.000 EMSZC14
581827 2012-11-08 09:00:06.000 EMSZC14
581827 2012-11-08 09:04:39.000 EMSZC49 Reset row number to 1
581827 2012-11-08 09:05:04.000 EMSZC49
581827 2012-11-08 09:06:32.000 EMSZC49
581827 2012-11-08 09:12:03.000 EMSZC49
581827 2012-11-08 09:12:38.000 EMSZC49
581827 2012-11-08 09:14:18.000 EMSZC49
581827 2012-11-08 09:17:35.000 EMSZC14 Reset row number to 1
-- my new rows for example of problem
581829 2012-11-08 09:12:03.000 EMSZC14 1
581829 2012-11-08 09:12:38.000 EMSZC49 1
581829 2012-11-08 09:14:18.000 EMSZC49
581829 2012-11-08 09:17:35.000 EMSZC14 Reset row number to 1
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新的解决方案的引入custno打破了这一解决方案,它非常适合custno.
with C1 as
(
select
custno, moddate, who,
lag(who) over(order by moddate) as lag_who
from
chr
),
C2 as
(
select
custno, moddate, who,
sum(case when who = lag_who then 0 else 1 end)
over(order by moddate rows unbounded preceding) as change
from
C1
)
select
row_number() over(partition by change order by moddate) as RowID,
custno, moddate, who
from
C2
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我确信这只是处理多个 custno 的一个小调整,但这已经远远超出了我的能力,我设法让它适用于我的数据,但这纯粹是通过替换列和表名称。不幸的是,没有足够详细的理解来解决我遇到的问题。
我的数据看起来像
custno start_date value
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实际上完全相同。我希望每当“值”或“谁”发生变化时,行/排名为 1,无论该值/谁之前是否见过。这都是相对于a而言的custno。我确实看到了一个值/谁也可以返回到相同值的实例。同样,上面的解决方案很好地处理了“重复”......但对于一个custno
custno我想我只需要以某种方式在某个地方添加某种分组依据?只是不确定在哪里或如何
谢谢!
这是一个间隙和岛屿问题,我们可以在这里使用行数差异法:
WITH cte AS (
SELECT *,
ROW_NUMBER() OVER (PARTITION BY custno ORDER BY moddate) rn1,
ROW_NUMBER() OVER (PARTITION BY custno, who ORDER BY moddate) rn2
FROM chr
)
SELECT custno, moddate, who,
ROW_NUMBER() OVER (PARTITION BY custno, rn1 - rn2 ORDER BY moddate) rn
FROM cte
ORDER BY
custno,
moddate;
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为了解释此处使用的行号方法的差异,rn1根据上面显示的数据,每个客户只是从 1 开始按时间排序的序列。该rn2序列另外由 划分who。对于每个客户来说,和之间的差异始终具有相同的价值。正是由于这种差异,我们然后在整个表上获取行号来生成您真正想要看到的序列。rn1rn2
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