如何在SQL中识别连续日期组？

Question

我正在尝试编写一个识别日期组的函数,并测量组的大小.

到目前为止,我一直在用Python编写程序,但是我想把它转移到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 

应输出到新表中:

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,这也应该能够解释周末和公众假期 - 我的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).有人在工作,建议我尝试CTE.我是新手,所以我很感激任何人都可以提供的任何指导!谢谢.

Answer 1

您可以通过巧妙的窗口函数应用程序来完成此操作.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,在您的示例中只会是1,2,3,4,5.现在,从日期中获取差值,并为组提供一个常量值.

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.我还会添加一个组标识符(例如,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,我错过了关于平日和假期的部分.该解决方案无法解决该问题.

Answer 2

以下查询帐户是周末和假日.查询有一个规定包括即时假期,但为了使查询更清楚,我只是将假期具体化为实际的表.

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. */

查询:

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转移回两天.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,则计算为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天成为第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,您需要在出勤时插入它们,因此可以轻松确定连续性,然后从最终查询中删除它们.上述出席人数连续11次出席.

查询逻辑的详细解释如下:http://www.ienablemuch.com/2012/07/monitoring-perfect-attendance.html