如何确定 SQL 服务器 "finding islands" 中的连续日期 count/days（连续行）

Question

我有一个学生 table，我想知道他们 session/training 持续了多长时间。我想排除周末，但我想计算不包括周末的连续天数。 A class 有 Start Date 和 End Date，例如，学生 ID S1 a 可以在 1 月预订 class 然后在 2 月再次预订，我想知道 1 月预订和 2 月预订有多少天，同时排除周末。基本上，我正在通过学生 ID 寻找从开始日期到结束日期的连续日期，除了周末没有休息时间。

SELECT 
 [ID]
,[StartDate]
,[EndDate]
,[BookingDays] AS Consecutive_Booking
FROM StudentBooking

如果学生 (student classifications(Type)) 在过去 3 天内预订了 class 5 天或 2 次（开始日期到结束日期（周一至周五））个月他们是 Resident else Visitors。开始日期和结束日期仅记录为周一至周五。请注意学生 ID 1 有一个连续的日期，这应该被算作一个块。 (02/01/2018-12/01/2018) 第二块 22/01-26/01

我想在下面复制 table。

ID   StartDate  EndDate     Duration     Type
1   02/01/2018  05/01/2018              ==>Please Note have continous dates
1   08/01/2018  12/01/2018   9           Resident
1   22/01/2018  26/01/2018   5           Resident 
2   23/01/2018  26/01/2018   4           Visitor
3   29/01/2018  31/01/2018   3           Visitor

Answer 1

这是我对你的问题的解决方案。

在 CTE "comparison" 中，我将每条记录与该学生的这条记录和所有后续记录相结合。这样，我就有了一个连续训练块的可能起点（从连接的左侧）和这样一个块的可能结束点（从连接的右侧）。使用 "cross applies"，我计算了 2 个值：

从链的第一个可能间隔开始到最后一个可能间隔结束的工作日
链中可能的最后一个间隔中的工作日。

根据后面的值，使用 windows 函数，我根据可能的开始和结束间隔构建了运行工作日总数。您用 "SQL 2012" 标记了问题，因此应该可以使用此 window 函数。

在下一个 CTE ("sorting") 中，我将之前的结果限制为运行总数等于第一个开始日期和最后一个结束日期之间的工作日。这样，只剩下连续的块。然后以两种方式对它们进行编号：

具有相同 EndDate 的连续块按 StartDate 升序编号
具有相同 StartDate 的连续块按 EndDate 降序编号。

对于每个 EndDate，我想要最早的 StartDate，而对于这个 StartDate，我只想要最新的 EndDate，所以我在两个编号中都筛选为 1。在这里：

WITH
  comparison (ID, StartDate, EndDate, TotalDays, SumSingleDays) AS (
    SELECT bStart.ID, bStart.StartDate, bEnd.EndDate, Workdays.Total
      , SUM(Workdays.Single) OVER (
          PARTITION BY bStart.ID, bStart.StartDate 
          ORDER BY bEnd.StartDate
          ROWS UNBOUNDED PRECEDING)
    FROM StudentBookings bStart
      INNER JOIN StudentBookings bEnd 
        ON bStart.ID = bEnd.ID AND bStart.StartDate <= bEnd.StartDate
      CROSS APPLY (VALUES (
        DATEDIFF(day, 0, bStart.StartDate), 
        DATEDIFF(day, 0, bEnd.StartDate), 
        1+DATEDIFF(day, 0, bEnd.EndDate))
      ) d (s1, s2, e2)
      CROSS APPLY (VALUES (
        (d.e2 - d.s1) - (d.e2/7 - d.s1/7) - ((d.e2+1)/7 - (d.s1+1)/7),
        (d.e2 - d.s2) - (d.e2/7 - d.s2/7) - ((d.e2+1)/7 - (d.s2+1)/7))
      ) Workdays (Total, Single)
  ),
  sorting (ID, StartDate, EndDate, Duration, RowNumStart, RowNumEnd) AS (
    SELECT ID, StartDate, EndDate, TotalDays
      , ROW_NUMBER() OVER (PARTITION BY ID, EndDate ORDER BY StartDate)
      , ROW_NUMBER() OVER (PARTITION BY ID, StartDate ORDER BY EndDate DESC)
    FROM comparison
    WHERE TotalDays = SumSingleDays
  )
SELECT ID, StartDate, EndDate, Duration
  , CASE WHEN Duration >= 5 THEN 'Resident' ELSE 'Visitor' END AS [Type]
FROM sorting 
WHERE (RowNumStart = 1) 
  AND (RowNumEnd = 1)
ORDER BY ID, StartDate;

结果：

也许有一个更优雅的方法来解决这个问题，使用 Itzik Ben-Gan 的间隔打包解决方案，我会 post 当我想出来的时候。

已添加：

此外，我计算了所有预订块的预订数量，并按学生 (ID) 求和，最终做出 "Resident" 决定。第一个 CTE（比较）的预订仅限于最近 3 个月：

WITH
  comparison (ID, StartDate, EndDate, TotalDays, CountBookings, SumSingleDays) AS (
    SELECT bStart.ID, bStart.StartDate, bEnd.EndDate, Workdays.Total
      , COUNT(Workdays.Single) OVER (
          PARTITION BY bStart.ID, bStart.StartDate 
          ORDER BY bEnd.StartDate
          ROWS UNBOUNDED PRECEDING)
      , SUM(Workdays.Single) OVER (
          PARTITION BY bStart.ID, bStart.StartDate 
          ORDER BY bEnd.StartDate
          ROWS UNBOUNDED PRECEDING)
    FROM StudentBookings bStart
      INNER JOIN StudentBookings bEnd 
        ON bStart.ID = bEnd.ID AND bStart.StartDate <= bEnd.StartDate
      CROSS APPLY (VALUES (
        DATEDIFF(day, 0, bStart.StartDate), 
        DATEDIFF(day, 0, bEnd.StartDate), 
        1+DATEDIFF(day, 0, bEnd.EndDate))
      ) d (s1, s2, e2)
      CROSS APPLY (VALUES (
        (d.e2 - d.s1) - (d.e2/7 - d.s1/7) - ((d.e2+1)/7 - (d.s1+1)/7),
        (d.e2 - d.s2) - (d.e2/7 - d.s2/7) - ((d.e2+1)/7 - (d.s2+1)/7))
      ) Workdays (Total, Single)
    WHERE bStart.StartDate >= DATEADD(month, -3, GETDATE())
  ),
  sorting (ID, StartDate, EndDate, Duration, CountBookings, RowNumStart, RowNumEnd) AS (
    SELECT ID, StartDate, EndDate, TotalDays, CountBookings
      , ROW_NUMBER() OVER (PARTITION BY ID, EndDate ORDER BY StartDate)
      , ROW_NUMBER() OVER (PARTITION BY ID, StartDate ORDER BY EndDate DESC)
    FROM comparison
    WHERE TotalDays = SumSingleDays
  ),
 counting (ID, StartDate, EndDate, Duration, Bookings) AS (
  SELECT ID, StartDate, EndDate, Duration
    , SUM(CountBookings) OVER (PARTITION BY ID)
  FROM sorting WHERE (RowNumStart = 1) AND (RowNumEnd = 1)
)
SELECT ID, StartDate, EndDate, Duration, Bookings
  , CASE 
      WHEN Duration >= 5 OR Bookings >= 2 THEN 'Resident' ELSE 'Visitor'
    END AS [Type]
FROM counting
ORDER BY ID, StartDate;

过滤Classe参考：

将从 bStart table 引用中获取并过滤 Class 引用。为了能够将此字段添加到最终查询中，还必须使用它来加入 bEndtable 参考，因此只有具有相同 Class参考值的预订间隔才会被连接块：

WITH
  comparison (ID, ClassReference, StartDate, EndDate, TotalDays, CountBookings, SumSingleDays) AS (
    SELECT bStart.ID, bStart.ClassReference, bStart.StartDate, bEnd.EndDate, Workdays.Total
      , COUNT(Workdays.Single) OVER (
          PARTITION BY bStart.ID, bStart.StartDate 
          ORDER BY bEnd.StartDate
          ROWS UNBOUNDED PRECEDING)
      , SUM(Workdays.Single) OVER (
          PARTITION BY bStart.ID, bStart.StartDate 
          ORDER BY bEnd.StartDate
          ROWS UNBOUNDED PRECEDING)
    FROM StudentBookings bStart
      INNER JOIN StudentBookings bEnd 
        ON bStart.ID = bEnd.ID AND bStart.StartDate <= bEnd.StartDate
       AND bStart.ClassReference = bEnd.ClassReference
      CROSS APPLY (VALUES (
        DATEDIFF(day, 0, bStart.StartDate), 
        DATEDIFF(day, 0, bEnd.StartDate), 
        1+DATEDIFF(day, 0, bEnd.EndDate))
      ) d (s1, s2, e2)
      CROSS APPLY (VALUES (
        (d.e2 - d.s1) - (d.e2/7 - d.s1/7) - ((d.e2+1)/7 - (d.s1+1)/7),
        (d.e2 - d.s2) - (d.e2/7 - d.s2/7) - ((d.e2+1)/7 - (d.s2+1)/7))
      ) Workdays (Total, Single)
    WHERE bStart.StartDate >= DATEADD(month, -3, GETDATE())
      AND bStart.ClassReference IN (N'C1', N'C2')
  ),
  sorting (ID, ClassReference, StartDate, EndDate, Duration, CountBookings, RowNumStart, RowNumEnd) AS (
    SELECT ID, ClassReference, StartDate, EndDate, TotalDays, CountBookings
      , ROW_NUMBER() OVER (PARTITION BY ID, ClassReference, EndDate ORDER BY StartDate)
      , ROW_NUMBER() OVER (PARTITION BY ID, ClassReference, StartDate ORDER BY EndDate DESC)
    FROM comparison
    WHERE TotalDays = SumSingleDays
  ),
  counting (ID, ClassReference, StartDate, EndDate, Duration, Bookings) AS (
    SELECT ID, ClassReference, StartDate, EndDate, Duration
      , SUM(CountBookings) OVER (PARTITION BY ID, ClassReference)
    FROM sorting WHERE (RowNumStart = 1) AND (RowNumEnd = 1)
  )
SELECT ID, ClassReference, StartDate, EndDate, Duration, Bookings
  , CASE 
      WHEN Duration >= 5 OR Bookings >= 2 THEN 'Resident' ELSE 'Visitor'
    END AS [Type]
FROM counting
ORDER BY ID, StartDate;

使用此数据进行测试：

使用最近 12 个月的过滤器，查询 returns:

所以学生 1 在 class C2 中是 "Resident"，但在 Class C1 中是访客。

如何确定 SQL 服务器 "finding islands" 中的连续日期 count/days（连续行）

How to Determine consecutive date count/days in SQL Server "finding islands" (consecutive rows)

sql

tsql

sql-server

sql-server-2012-express