2013-06-21 74 views
0

SQL MIN和MAX的關係,我從一個SQL表,可以使用下面的代碼檢索列 -在SELECT CASE

MAX(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN bd.ID ELSE NULL END) "MondayHourlyAM",     
      MAX(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN s.SchoolName ELSE NULL END) "MondayHourlyAMSchoolName", 
      MAX(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN REPLACE(REPLACE(LTRIM(RIGHT(CONVERT(VARCHAR, bd.StartTime, 100), 7)) + '-' + LTRIM(RIGHT(CONVERT(VARCHAR(20), bd.EndTime, 100), 7)), 'AM',''), 'PM', '') ELSE NULL END) "MondayHourlyAMTimes", 
      MAX(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN bd.TotalChargeAmount ELSE NULL END) "MondayHourlyAMTotalChargeAmount", 
      MAX(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN bd.TotalPayAmount ELSE NULL END) "MondayHourlyAMTotalPayAmount", 
      MAX(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN bd.BandBookedAt ELSE NULL END) "MondayHourlyAMBandBookedAt", 

      MIN(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN bd.ID ELSE NULL END) "MondayHourlyAM2",      
      MIN(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN s.SchoolName ELSE NULL END) "MondayHourlyAM2SchoolName", 
      MIN(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN REPLACE(REPLACE(LTRIM(RIGHT(CONVERT(VARCHAR, bd.StartTime, 100), 7)) + '-' + LTRIM(RIGHT(CONVERT(VARCHAR(20), bd.EndTime, 100), 7)), 'AM',''), 'PM', '') ELSE NULL END) "MondayHourlyAM2Times", 
      MIN(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN bd.TotalChargeAmount ELSE NULL END) "MondayHourlyAM2TotalChargeAmount", 
      MIN(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN bd.TotalPayAmount ELSE NULL END) "MondayHourlyAM2TotalPayAmount", 
      MIN(CASE WHEN bd.DayText = 'Monday' and bd.BookingDuration = 3 and CONVERT(time(0), bd.StartTime) < CONVERT(time(0), '12:00:00') AND bd.NoOfHOurs < 5.5 and s.PrimarySchool = 1 THEN bd.BandBookedAt ELSE NULL END) "MondayHourlyAM2BandBookedAt", 

運行此代碼時,我得到了2個預訂,但返回的充電/收費率爲其他預訂。即 -

應該是 預訂1-電荷£20,£付費10 預訂2-充電£100,付費£50

但即時得到 預訂1 - 充電£100,付費£50 預訂2 - 收費£20,支付£10

我以爲這是因爲它使用MAX和MIN返回最高和最低值。我如何將MAX和MIN值返回到「HourlyAM」列?

希望我解釋得很清楚!由於

+0

給出幾乎所有的case語句具有相同的邏輯,您應該將common子句從case中移出並放入查詢的主'where'子句中。 –

+0

這是一個更大的查詢的一小部分。 (IE:AM,PM,FullDay,AMHourly,PMHourly,FullDayHourly)每週的每一天都有這些段落中的6段,類似但持續時間不同。我意識到它不是最有效的方式,但它似乎運作良好,而且速度很快。因此,我認爲我不能將共同的條款移到主要的where子句中。乾杯 – dynamicuser

回答

2

我不打算嘗試通過您的標準沒有更好的瞭解您的數據涉水,但我認爲你會使用ROW_NUMBER()得到更好的服務:

SELECT * 
FROM (SELECT *, ROW_NUMBER() OVER (ORDER BY MondayHourlyAMTimes DESC) 'Booking1' 
       , ROW_NUMBER() OVER (ORDER BY MondayHourlyAMTimes) 'Booking2' 
     FROM Table 
     WHERE common criteria 
    ) 
WHERE Booking1 = 1 OR Booking2 = 1 

的想法在這裏是,您可以根據一個字段爲每個滿足條件的行分配排名,然後根據該排名返回完整的行。