我正在尝试编写一个识别日期组并测量组大小的函数。
到目前为止，我一直在 Python 中按程序执行此操作，但我想将其移至 SQL 中。
例如，列表
Bill 01/01/2011
Bill 02/01/2011
Bill 03/01/2011
Bill 05/01/2011
Bill 07/01/2011
应该输出到一个新表中：
Bill 01/01/2011 3
Bill 02/01/2011 3
Bill 03/01/2011 3
Bill 05/01/2011 1
Bill 07/01/2011 1
理想情况下，这也应该能够考虑周末和公共假期 - 我表中的日期将是周一至周五（我想我可以通过制作一个新的工作日表并按顺序编号来解决这个问题）。工作中有人建议我尝试 CTE。我对此很陌生，所以我很感激任何人可以提供的任何指导！谢谢。
您可以通过巧妙地应用窗口函数来做到这一点。考虑以下：
select name, date, row_number() over (partition by name order by date)
from t
这会添加一个行号，在您的示例中它只是 1、2、3、4、5。现在，取日期的差值，您就有了该组的常量值。
select name, date,
dateadd(d, - row_number() over (partition by name order by date), date) as val
from t
最后，您需要按顺序排列的组数。我还将添加一个组标识符（例如，以区分最后两个）。
select name, date,
count(*) over (partition by name, val) as NumInSeq,
dense_rank() over (partition by name order by val) as SeqID
from (select name, date,
dateadd(d, - row_number() over (partition by name order by date), date) as val
from t
) t
不知何故，我错过了关于工作日和节假日的部分。此解决方案不能解决该问题。
以下查询帐户周末和节假日。该查询提供了即时包含假期的规定，但为了使查询更清晰，我只是将假期具体化到一个实际的表中。
CREATE TABLE tx
(n varchar(4), d date);
INSERT INTO tx
(n, d)
VALUES
('Bill', '2006-12-29'), -- Friday
-- 2006-12-30 is Saturday
-- 2006-12-31 is Sunday
-- 2007-01-01 is New Year's Holiday
('Bill', '2007-01-02'), -- Tuesday
('Bill', '2007-01-03'), -- Wednesday
('Bill', '2007-01-04'), -- Thursday
('Bill', '2007-01-05'), -- Friday
-- 2007-01-06 is Saturday
-- 2007-01-07 is Sunday
('Bill', '2007-01-08'), -- Monday
('Bill', '2007-01-09'), -- Tuesday
('Bill', '2012-07-09'), -- Monday
('Bill', '2012-07-10'), -- Tuesday
('Bill', '2012-07-11'); -- Wednesday
create table holiday(d date);
insert into holiday(d) values
('2007-01-01');
/* query should return 7 consecutive good
attendance(from December 29 2006 to January 9 2007) */
/* and 3 consecutive attendance from July 7 2012 to July 11 2012. */
询问：
with first_date as
(
-- get the monday of the earliest date
select dateadd( ww, datediff(ww,0,min(d)), 0 ) as first_date
from tx
)
,shifted as
(
select
tx.n, tx.d,
diff = datediff(day, fd.first_date, tx.d)
- (datediff(day, fd.first_date, tx.d)/7 * 2)
from tx
cross join first_date fd
union
select
xxx.n, h.d,
diff = datediff(day, fd.first_date, h.d)
- (datediff(day, fd.first_date, h.d)/7 * 2)
from holiday h
cross join first_date fd
cross join (select distinct n from tx) as xxx
)
,grouped as
(
select *, grp = diff - row_number() over(partition by n order by d)
from shifted
)
select
d, n, dense_rank() over (partition by n order by grp) as nth_streak
,count(*) over (partition by n, grp) as streak
from grouped
where d not in (select d from holiday) -- remove the holidays
输出：
| D | N | NTH_STREAK | STREAK |
-------------------------------------------
| 2006-12-29 | Bill | 1 | 7 |
| 2007-01-02 | Bill | 1 | 7 |
| 2007-01-03 | Bill | 1 | 7 |
| 2007-01-04 | Bill | 1 | 7 |
| 2007-01-05 | Bill | 1 | 7 |
| 2007-01-08 | Bill | 1 | 7 |
| 2007-01-09 | Bill | 1 | 7 |
| 2012-07-09 | Bill | 2 | 3 |
| 2012-07-10 | Bill | 2 | 3 |
| 2012-07-11 | Bill | 2 | 3 |
现场测试：http ://www.sqlfiddle.com/#!3/815c5/1
查询的主要逻辑是将所有日期向后移动两天。这是通过将日期除以 7 并将其乘以 2，然后从原始数字中减去它来完成的。例如，如果给定日期是 15 日，这将计算为 15/7 * 2 == 4；然后从原始数字中减去 4，15 - 4 == 11。15 将成为第 11 天。同样，第 8 天变成第 6 天；8 - (8/7 * 2) == 6。
Weekends are not in attendance(e.g. 6,7,13,14)
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15
将计算应用于所有工作日数字将产生以下值：
1 2 3 4 5
6 7 8 9 10
11
对于假期，您需要在出勤时将它们插入，以便可以轻松确定连续性，然后将它们从最终查询中删除。以上出勤产生连续11次良好出勤。
查询逻辑的详细解释在这里：http ://www.ienablemuch.com/2012/07/monitoring-perfect-attendance.html