MIN的替代选择一个以上的行

Question

下面的SQL查询应该显示教师的可用性。 有3种可能的预订类型-AM，PM或全天。 如果有AM预订，则单元格中的文本应显示PM，如果有PM预订，则应显示AM，如果有全天预订，或者AM和PM预订都应显示“ xxx”。

当有1 AM，1 PM或1全天预订时，这一切都很好，但是如果同一天有1 AM和1 PM，则不会显示XXX，因为我正在使用MIN（）。 我如何才能在一天之内而不是最小的一天来评估所有预订？ （您可以看到我试图在代码的星期四部分显示我的意思。

SQL：

with CTE_D as 
( 
   SELECT 
      DATEADD(ww, DATEDIFF(ww,0,GETDATE()), 0) as BookingDate 

   UNION ALL

   SELECT
      DATEADD(day, 1, BookingDate) 
   FROM
      CTE_D 
   WHERE
      BookingDate < DATEADD(ww, DATEDIFF(ww,0,GETDATE()), 6) 
)
SELECT 
    t.ID, t.Firstname, 
    t.Surname, tb.Band, t.Telephone, t.Mobile, t.Teacher, t.TeacherAssistant, t.PrimarySchool, t.SecondarySchool,
    MIN(CASE WHEN bd.DayText = 'Monday' AND bd.BookingDuration = 0 THEN 'PM' ELSE 'Full Day' END) "Monday", 
    MIN(CASE WHEN bd.DayText = 'Tuesday' AND bd.BookingDuration = 0 THEN 'PM' ELSE 'Full Day' END) "Tuesday", 
    MIN(CASE WHEN bd.DayText = 'Wednesday' AND bd.BookingDuration = 0 THEN 'PM' ELSE 'Full Day' END) "Wednesday", 
    MIN(CASE WHEN bd.DayText = 'Thursday' AND bd.BookingDuration = 0 THEN 'PM' WHEN bd.DayText = 'Thursday' AND bd.BookingDuration = 1 THEN 'AM' WHEN bd.DayText = 'Thursday' AND bd.BookingDuration = 2 or (bd.BookingDuration = 1 and bd.BookingDuration = 0) THEN 'xxx' END) "Thursday", 
    MIN(CASE WHEN bd.DayText = 'Friday' AND bd.BookingDuration = 0 THEN 'PM' ELSE 'Full Day' END) "Friday",
    Notes 
        FROM Teachers t 
        cross join CTE_D d 
        left join BookingDays bd 
            on t.ID = bd.TeacherID and 
               bd.BookingDate = d.BookingDate 
        left join BookingDurations bds 
            on bd.BookingDuration = bds.ID 
        left join TeacherBands tb on t.Band = tb.ID
        WHERE t.Active = 0 and (t.Status = 0 or t.Status = 1) and (bd.Status = 0 or bd.Status IS NULL) and PrimarySchool = 1
        GROUP BY Firstname, Surname, t.Telephone, t.Mobile, t.Notes, tb.Band, t.ID, t.Teacher, t.TeacherAssistant, t.PrimarySchool, t.SecondarySchool, t.Nursery, t.Reception, t.Year1, t.Year2, t.Year3, t.Year4, t.Year5, t.Year6, t.Year7, t.Year8, t.Year9, t.Year10, t.Year11, t.ALevel
        ORDER BY Surname, Firstname ASC

表格栏：

ID | Firstname | Surname | Band | Telephone | Mobile | Teacher | Teacher Assistant | PrimarySchool | SecondarySchool | KeyStage | Mon | Tues | Wed | Thurs | Fri | Notes

谢谢，马特

Answer 1

我认为问题在于您正在尝试同时执行两个需要分开的操作（请检查工作日，并检查预订持续时间）。 为此，我假设您的预订时长为：

• 0-下午
• 1-上午
• 2-全天

如果这不正确，则需要调整我的查询范围，但这应该可以使您走上正确的道路。 我还假设预订表中没有记录意味着不可用：

SET LANGUAGE British;

WITH Bookings AS
(   SELECT  BookingDays.TeacherID,
            [WeekDay] = DATENAME(WEEKDAY, BookingDate),
            [Status] = CASE WHEN BookingDuration = 0 THEN 'PM' WHEN BookingDuration = 1 THEN 'AM' ELSE 'Full Day' END
    FROM    BookingDays
), PivotedBookings AS
(   SELECT  *
    FROM    Bookings
            PIVOT
            (   MAX([Status])
                FOR [WeekDay] IN ([Monday], [Tuesday], [Wednesday], [Thursday], [Friday])
            ) pvt
)
SELECT  t.ID, 
        t.Firstname, 
        t.Surname, 
        tb.Band, 
        t.Telephone, 
        t.Mobile, 
        t.Teacher, 
        t.TeacherAssistant, 
        t.PrimarySchool, 
        t.SecondarySchool,
        Monday = COALESCE(pb.Monday, 'Not Available'),
        Tuesday = COALESCE(pb.Tuesday, 'Not Available'),
        Wednesday = COALESCE(pb.Wednesday, 'Not Available'),
        Thursday = COALESCE(pb.Thursday, 'Not Available'),
        Friday = COALESCE(pb.Friday, 'Not Available'),
        t.Notes
FROM    Teachers t
        LEFT JOIN PivotedBookings pb
            ON pb.TeacherID = t.ID
        LEFT JOIN TeacherBands tb
            ON tb.ID = t.Band;

减少SQL Fiddle上的示例

---

附录

阅读此行：

WHEN bd.DayText = 'Thursday' AND bd.BookingDuration = 2 or (bd.BookingDuration = 1 and bd.BookingDuration = 0)

这使我相信全天有2种可能性，预订持续时间有2种，或者表中同一位老师和日期的AM和PM条目。 在这种情况下，有必要将数据轮换两次，因此您的第一个CTE变为：

WITH Bookings AS
(   SELECT  BookingDays.TeacherID,
            [WeekDay] = DATENAME(WEEKDAY, BookingDate),
            [Status] = CASE WHEN [2] > 0 THEN 'Full Day'
                            WHEN [0] > 0 AND [1] > 0 THEN 'xxx'
                            WHEN [0] > 0 THEN 'PM'
                            WHEN [1] > 0 THEN 'AM'
                        END

    FROM    (   SELECT  TeacherID, BookingDate, BookingDuration, [X] = 1
                FROM    BookingDays
            ) BookingDays
            PIVOT
            (   SUM(X)
                FOR BookingDuration IN ([0], [1], [2])
            ) pvt
)

更新了SQL Fiddle上的示例（教师1注意周四）