如何在SQL中获取连续天数并在不再连续时中断

Question

我正在尝试创建一个查询，该查询计算员工的连续工作天数。如果连续两天都有休息，它应该会重置，但我无法正常工作。任何想法将不胜感激！

这是我的查询。我正在尝试使用row_number函数，但是我不确定这是否正确：

WITH CTE AS (
      select distinct TOT.EMPLOYEEID, TOT.DATE
      from TOTALS TOT
      where TOT.EMPLOYEEID IN ('020576','1200823') and
            TOT.TIMEINSECONDS >= 14400 and
            TOT.DATE >= '2019-01-01'
     )
SELECT CTE.*,
       ROW_NUMBER() OVER (PARTITION BY EMPLOYEEID ORDER BY DATE) AS CONSECUTIVEDAYS
FROM CTE
ORDER BY EMPLOYEEID, DATE;

这里是结果：

EMPLOYEEID        DATE        CONSECUTIVEDAYS
020576        2019-01-01 00:00:00.000        1
020576        2019-01-02 00:00:00.000        2
020576        2019-01-03 00:00:00.000        3
020576        2019-01-04 00:00:00.000        4
020576        2019-01-07 00:00:00.000        5 <---- THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.

020576        2019-01-08 00:00:00.000        6
020576        2019-01-09 00:00:00.000        7
020576        2019-01-10 00:00:00.000        8
020576        2019-01-11 00:00:00.000        9
020576        2019-01-14 00:00:00.000        10 <---THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.

020576        2019-01-15 00:00:00.000        11
020576        2019-01-16 00:00:00.000        12
020576        2019-01-17 00:00:00.000        13
020576        2019-01-21 00:00:00.000        14 <---THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.

020576        2019-01-22 00:00:00.000        15
020576        2019-01-23 00:00:00.000        16
020576        2019-01-24 00:00:00.000        17
020576        2019-01-25 00:00:00.000        18
020576        2019-01-28 00:00:00.000        19 <---THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.

020576        2019-01-29 00:00:00.000        20
020576        2019-01-30 00:00:00.000        21
020576        2019-01-31 00:00:00.000        22
1200823        2019-01-01 00:00:00.000        1
1200823        2019-01-02 00:00:00.000        2
1200823        2019-01-03 00:00:00.000        3
1200823        2019-01-04 00:00:00.000        4
1200823        2019-01-07 00:00:00.000        5 <---THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.
1200823        2019-01-08 00:00:00.000        6
1200823        2019-01-09 00:00:00.000        7
1200823        2019-01-10 00:00:00.000        8
1200823        2019-01-11 00:00:00.000        9
1200823        2019-01-14 00:00:00.000        10 <---THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.
1200823        2019-01-15 00:00:00.000        11
1200823        2019-01-16 00:00:00.000        12
1200823        2019-01-18 00:00:00.000        13 <---THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.
1200823        2019-01-21 00:00:00.000        14 <---THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.
1200823        2019-01-22 00:00:00.000        15
1200823        2019-01-23 00:00:00.000        16
1200823        2019-01-24 00:00:00.000        17
1200823        2019-01-25 00:00:00.000        18
1200823        2019-01-28 00:00:00.000        19 <---THIS SHOULD BE 1. THE COUNT SHOULD RESET BECAUSE THE DATES ARE NO LONGER CONSECUTIVE.
1200823        2019-01-29 00:00:00.000        20
1200823        2019-01-30 00:00:00.000        21

Answer 1

您可以用不同的方式标识相邻行号的组。一种方法是使用LAG()并确定组的起始位置。然后，累加和定义了组。

最后一步是对每个组进行ROW_NUMBER()，以获得所需的编号：

WITH CTE AS (
      select distinct TOT.EMPLOYEEID, TOT.DATE
      from TOTALS TOT
      where TOT.EMPLOYEEID IN ('020576','1200823') and
            TOT.TIMEINSECONDS >= 14400 and
            TOT.DATE >= '2019-01-01'
     )
SELECT t.*,
        ROW_NUMBER() OVER PARTITION BY EMPLOYEEID, GRP ORDER BY Date) as seqnum
FROM (SELECT CTE.*,
             SUM(CASE WHEN prev_date < date - interval '1 day' THEN 1 ELSE 0 END) OVER (PARTITION BY EMPLOYEEID ORDER BY DATE) as grp
             ROW_NUMBER() OVER (PARTITION BY EMPLOYEEID ORDER BY DATE) AS CONSECUTIVEDAYS
      FROM (SELECT CTE.*,
                   LAG(DATE) OVER (PARTITION BY EMPLOYEEID ORDER BY DATE) as prev_date
            FROM CTE
           ) t
     ) t
ORDER BY EMPLOYEEID, DATE;

请注意，这使用通用日期函数。日期处理众所周知是特定于数据库的。它应该很容易适应您实际使用的数据库。