从每日粒度的事实表计算日期间隔答案

【问题标题】：Calculating date intervals from a daily-grained fact table从每日粒度的事实表计算日期间隔
【发布时间】：2018-08-09 14:12:15
【问题描述】：

我有一些转换后得到的学生缺勤数据。数据是一天一天的：

WITH datasample AS (
    SELECT 1 AS StudentID, 20180101 AS DateID, 0 AS AbsentToday, 0 AS AbsentYesterday UNION ALL
    SELECT 1, 20180102, 1, 0 UNION ALL
    SELECT 1, 20180103, 1, 1 UNION ALL
    SELECT 1, 20180104, 1, 1 UNION ALL
    SELECT 1, 20180105, 1, 1 UNION ALL
    SELECT 1, 20180106, 0, 1 UNION ALL
    SELECT 2, 20180101, 0, 0 UNION ALL
    SELECT 2, 20180102, 1, 0 UNION ALL
    SELECT 2, 20180103, 1, 1 UNION ALL
    SELECT 2, 20180104, 0, 1 UNION ALL
    SELECT 2, 20180105, 1, 0 UNION ALL
    SELECT 2, 20180106, 1, 1 UNION ALL
    SELECT 2, 20180107, 0, 1
)
SELECT *
FROM datasample
ORDER BY StudentID, DateID

我需要添加一个列 (AbsencePeriodInMonth) 来计算学生在该月的缺勤时间。例如，StudentID=1 在一个月内连续一个时段缺席，StudentID=2 有两个时段，如下所示：

StudentID DateID    AbsentToday AbsentYesterday AbsencePeriodInMonth
1         20180101  0           0               0
1         20180102  1           0               1
1         20180103  1           1               1
1         20180104  1           1               1
1         20180105  1           1               1
1         20180106  0           1               0
2         20180101  0           0               0
2         20180102  1           0               1
2         20180103  1           1               1
2         20180104  0           1               0
2         20180105  1           0               2
2         20180106  1           1               2
2         20180107  0           1               0

我的目标实际上是计算事实表中每一天之前的连续缺席天数，我想如果我得到 AbsencePeriodInMonth 列，我可以通过在 * 之后将其添加到我的查询中来做到这一点：

,CASE WHEN AbsentToday = 1 THEN DENSE_RANK() OVER(PARTITION BY StudentID, AbsencePeriodInMonth ORDER BY DateID)
           ELSE 0
     END AS DaysAbsent

知道如何添加 AbsencePeriodInMonth 或以其他方式计算连续缺勤天数吗？

【问题讨论】：

标签： sql sql-server

【解决方案1】：

您可以通过事先计算 0 的数量来识别每个周期。然后您可以使用dense_rank() 枚举它们。

select ds.*,
       (case when absenttoday = 1 then dense_rank() over (partition by studentid order by grp)
             else 0
        end) as AbsencePeriodInMonth
from (select ds.*, sum(case when absenttoday = 0 then 1 else 0 end) over (partition by studentid order by dateid) as grp
      from datasample ds
     ) ds
order by StudentID, DateID;

Here 是一个 SQL Fiddle。

【讨论】：

似乎使用dense_rank() 是多余的。
@AjayGupta 。 . .如果“缺勤期”将在每个新期间被枚举为 1，则需要它。
不等于grp
谢谢，grp 会随着不缺席的每一天而增加，但它可以识别不同的缺席时段，这正是我所需要的

【解决方案2】：

使用Recursive CTE 和Dense_Rank

WITH datasample AS (
    SELECT 1 AS StudentID, 20180101 AS DateID, 0 AS AbsentToday, 0 AS AbsentYesterday UNION ALL
    SELECT 1, 20180102, 1, 0 UNION ALL
    SELECT 1, 20180103, 1, 1 UNION ALL
    SELECT 1, 20180104, 1, 1 UNION ALL
    SELECT 1, 20180105, 1, 1 UNION ALL
    SELECT 1, 20180106, 0, 1 UNION ALL
    SELECT 2, 20180101, 0, 0 UNION ALL
    SELECT 2, 20180102, 1, 0 UNION ALL
    SELECT 2, 20180103, 1, 1 UNION ALL
    SELECT 2, 20180104, 0, 1 UNION ALL
    SELECT 2, 20180105, 1, 0 UNION ALL
    SELECT 2, 20180106, 1, 1 UNION ALL
    SELECT 2, 20180107, 0, 1
), cte as
(Select *,DateID as dd 
from datasample 
where AbsentToday = 1 and AbsentYesterday = 0

union all

Select d.*, c.dd 
from datasample d
join cte c
on d.StudentID = c.StudentID and d.DateID = c.DateID + 1 
    where d.AbsentToday = 1
), cte1 as
(
Select *, DENSE_RANK() over (partition by StudentId order by dd) as de 
from cte
)
Select d.*, IsNull(c.de,0) as AbsencePeriodInMonth
from cte1 c 
right join datasample d
on d.StudentID = c.StudentID and c.DateID = d.DateID
order  by d.StudentID, d.DateID

【讨论】：