使用SQL Server 2012，我正在尝试创建一个查询，为我提供气候数据库中排名前10位最长的潮湿（或干燥）时段。
我的临时表提供以下数据输出：
select monthid as [id], date, rain_today
from #raindays
order by monthid asc, date asc
输出：
id date rain_today
-------------------------------
1 24 Dec 2014 2.4
1 25 Dec 2014 0
1 26 Dec 2014 8.7
1 27 Dec 2014 1.8
1 28 Dec 2014 0.3
1 29 Dec 2014 0
1 30 Dec 2014 0
1 31 Dec 2014 0.3
2 01 Jan 2015 0.3
2 02 Jan 2015 0.3
2 03 Jan 2015 18.3
2 04 Jan 2015 0.3
等。等
我想返回一个排名表，该表将计算rain_today>的时间段。 0，（或rain_today = 0）即：
Rank Start_Date End_Date Wet Period
----------------------------------------
1 31 Dec 2014 04 Jan 2015 5
2 26 Dec 2014 28 Dec 2014 3
...
我从审核其他类似查询得到的最接近的是（这是干天）：
select
#raindays.monthid as id,
min(#raindays.date) as [FirstDryDay],
max(#raindays.date) as [LatestDryDay],
count(*) as countdays
from
(select
monthid,
coalesce(max(case
when rain_today > '0'
then #raindays.date end), '19000101') as latestdry
from
#raindays
group by
monthid) g
join
#raindays on #raindays.monthid = g.monthid
and #raindays.date > g.latestdry
group by
#raindays.monthid
order by
countdays desc
输出：
id FirstDryDay LatestDryDay countdays
-----------------------------------------------
23 21 Oct 2016 31 Oct 2016 11
21 23 Aug 2016 31 Aug 2016 9
**15 23 Feb 2016 29 Feb 2016 7**
10 25 Sep 2015 30 Sep 2015 6
8 28 Jul 2015 31 Jul 2015 4
24 28 Nov 2016 30 Nov 2016 3
29 29 Apr 2017 30 Apr 2017 2
30 30 May 2017 31 May 2017 2
31 29 Jun 2017 30 Jun 2017 2
20 30 Jul 2016 31 Jul 2016 2
7 29 Jun 2015 30 Jun 2015 2
5 30 Apr 2015 30 Apr 2015 1
11 31 Oct 2015 31 Oct 2015 1
17 30 Apr 2016 30 Apr 2016 1
22 30 Sep 2016 30 Sep 2016 1
正如你所看到的，我真的不想按ID分组，因为我希望能够跨越不同的月份而且我错过了本月早些时候发生的其他时期。实际计数看起来正常，检查上面突出显示的时间段：
id date rain_today
15 22 Feb 2016 3.9
15 23 Feb 2016 0
15 24 Feb 2016 0
15 25 Feb 2016 0
15 26 Feb 2016 0
15 27 Feb 2016 0
15 28 Feb 2016 0
15 29 Feb 2016 0
16 01 Mar 2016 3
提前感谢您的帮助！
答案 0 :(得分：2)
这是你想要的吗？
IF OBJECT_ID('tempdb..#TestData', 'U') IS NOT NULL
DROP TABLE #TestData;
CREATE TABLE #TestData (
id INT NOT NULL ,
[Date] DATE NOT NULL,
Rain_Today DECIMAL(9,2) NOT NULL
);
INSERT #TestData (id, Date, Rain_Today) VALUES
(1, '24 Dec 2014', 2.4),
(1, '25 Dec 2014', 0),
(1, '26 Dec 2014', 8.7),
(1, '27 Dec 2014', 1.8),
(1, '28 Dec 2014', 0.3),
(1, '29 Dec 2014', 0),
(1, '30 Dec 2014', 0),
(1, '31 Dec 2014', 0.3),
(2, '01 Jan 2015', 0.3),
(2, '02 Jan 2015', 0.3),
(2, '03 Jan 2015', 18.3),
(2, '04 Jan 2015', 0.3);
--======================================
WITH
cte_AddRankGroup AS (
SELECT
td.id,
td.Date,
td.Rain_Today,
hr.HasRain,
RankGroup = DENSE_RANK() OVER (PARTITION BY td.id ORDER BY td.Date) -
DENSE_RANK() OVER (PARTITION BY td.id, hr.HasRain ORDER BY td.Date)
FROM
#TestData td
CROSS APPLY ( VALUES (IIF(td.Rain_Today = 0, 0, 1)) ) hr (HasRain)
)
SELECT
arg.id,
BegDate = MIN(arg.Date),
EndDate = MAX(arg.Date),
WetPeriod = IIF(arg.HasRain = 1, 'Wet', 'Dry'),
ConsecutiveDays = COUNT(1)
FROM
cte_AddRankGroup arg
GROUP BY
arg.id,
arg.HasRain,
arg.RankGroup
ORDER BY
arg.id,
MIN(arg.Date);
结果...
id BegDate EndDate WetPeriod ConsecutiveDays
----------- ---------- ---------- --------- ---------------
1 2014-12-24 2014-12-24 Wet 1
1 2014-12-25 2014-12-25 Dry 1
1 2014-12-26 2014-12-28 Wet 3
1 2014-12-29 2014-12-30 Dry 2
1 2014-12-31 2014-12-31 Wet 1
2 2015-01-01 2015-01-04 Wet 4
编辑：代码版本使用CASE表达式代替IIF ...
--======================================
WITH
cte_AddRankGroup AS (
SELECT
td.id,
td.Date,
td.Rain_Today,
hr.HasRain,
RankGroup = DENSE_RANK() OVER (PARTITION BY td.id ORDER BY td.Date) -
DENSE_RANK() OVER (PARTITION BY td.id, hr.HasRain ORDER BY td.Date)
FROM
#TestData td
CROSS APPLY ( VALUES (CASE WHEN td.Rain_Today = 0 THEN 0 ELSE 1 END) ) hr (HasRain)
)
SELECT top 10
arg.id,
BegDate = MIN(arg.Date),
EndDate = MAX(arg.Date),
WetPeriod = CASE WHEN arg.HasRain = 1 THEN 'Wet' ELSE 'Dry' END,
ConsecutiveDays = COUNT(1)
FROM
cte_AddRankGroup arg
WHERE
arg.HasRain = '0' -- Top 10 Dry
--arg.HasRain = '1' -- Top 10 Wet
GROUP BY
arg.id,
arg.HasRain,
arg.RankGroup
ORDER BY
ConsecutiveDays desc, MIN(arg.Date);
修改原始脚本以按每个句点类型生成前10名，这是我的最终目标（输出来自完整数据集）：
id BegDate EndDate WetPeriod ConsecutiveDays
31 10 Jun 2017 26 Jun 2017 Dry 17
4 02 Mar 2015 14 Mar 2015 Dry 13
5 12 Apr 2015 24 Apr 2015 Dry 13
20 15 Jul 2016 26 Jul 2016 Dry 12
29 01 Apr 2017 11 Apr 2017 Dry 11
26 17 Jan 2017 27 Jan 2017 Dry 11
23 21 Oct 2016 31 Oct 2016 Dry 11
25 01 Dec 2016 09 Dec 2016 Dry 9
21 10 Aug 2016 18 Aug 2016 Dry 9
21 23 Aug 2016 31 Aug 2016 Dry 9
答案 1 :(得分：0)
这个问题可以通过递归方式解决：
-- this variable is needed to stop the recursion
declare @numrows int=(select count(1) from #raindays)
-- add a row number to the table creating a new table as "tabseq"
;WITH tabseq as (select row_number() over(order by date) as rownum, * from #raindays),
-- apply recursion to tabseq keeping a toggle running totals of wet and dry periods
CTE as
(
select *,
(case when rain_today=0 then 1 else 0 end) as dry,
(case when rain_today>0 then 1 else 0 end) as wet
from tabseq where rownum=1
union all
select s.*,
(case when s.rain_today=0 then cte.dry+1 else 0 end) as dry,
(case when s.rain_today>0 then cte.wet+1 else 0 end) as wet
from tabseq s
join cte on s.rownum=cte.rownum+1
where s.rownum<=@numrows
)
select * from cte
一旦您将表（cte）与干/湿蓄电池配合使用，您可以订购并选择它以满足您的输出要求。
请注意，这是假设连续几天在桌子上，如果有差距，那么在案例陈述中添加+1而不是在一方或另一方添加一个日期，取决于您如何考虑缺少日期（湿的还是干的。