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，我一直在使用Python进行程序化处理，但我想把它转移到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

应将

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 - 我的表格中的日期将是周一至周五（我想我可以通过制作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！感谢。

Answer 1

您可以通过巧妙的窗口功能应用程序来完成此操作。请考虑以下事项：

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，从日期中获取差异，并为该组保留一个常量值。

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，可以即时包含假期，但为了使查询更清晰，我只是将假期具体化为实际的表格。

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 |

实时测试：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，你需要在出席时插入它们，因此可以很容易地确定连续性，然后从最终查询中删除它们。上述出席率连续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

如何在SQL中识别连续日期组？

2 个答案: