獲取計數的總和與條件

所以我試圖讓具有條件（從查詢的其餘部分不同）的計數的總和獲取計數的總和與條件

這是我到目前爲止有：

SELECT studemo.ssid         AS [SSID], 
     stustat.graden        AS Grade, 
     studemo.lastname        AS [Last Name], 
     studemo.firstname       AS [First Name], 
     (SELECT Count(*) 
     WHERE attend.attendc IN ('E', 'G', 'H', 'I', 
            'J', 'L', 'M', 'P', 
            'Q', 'V', 'X')) AS [Days Absent], 
     attend.attendc        AS [Attendance Code], 
     attend.ddate         AS [Date] 
FROM track, 
     stustat, 
     stusched, 
     studemo, 
     attend 
WHERE studemo.suniq = stustat.suniq 
     AND attend.scduniq = stusched.scduniq 
     AND studemo.suniq = stusched.suniq 
     AND stustat.trkuniq = track.trkuniq 
     AND track.schoolc = '408' 
     AND track.schyear = '2013' 
     AND stustat.edate >= '08/21/2012' 
     AND stustat.xdate IS NULL 
ORDER BY [last name], 
      [first name], 
      attend.ddate

這會給我正確的信息，只是不是我想要得到它的方式。它會給我一行每一個出席代碼。我希望能做的就是加總從count（*）子查詢中獲得某些東西的次數。我嘗試以大約100種不同的方式求和（），但我無法得到它。有什麼建議麼？

來源

2013-05-10 m0ngr31

展示樣品輸入和輸出數據...它很難理解從這裏... – 2013-05-10 18:08:39

談隱加入 – Kermit 2013-05-10 18:09:53

誰像這樣寫代碼？ – 2013-05-10 18:14:38

嘗試更換行：

(select count(*) where attend.attendc IN ('E','G','H','I','J','L','M','P','Q','V','X')) AS [Days Absent],

與窗函數：

sum(case when attend.attendc IN ('E','G','H','I','J','L','M','P','Q','V','X') then 1 else 0 end) over() as [Days Absent]

也就是說天缺席的總數。我猜你想要這個SSID。在這種情況下，使用partition by：

sum(case when attend.attendc IN ('E','G','H','I','J','L','M','P','Q','V','X') then 1 else 0 end) 
      over (partition by SSID) as [Days Absent]

來源

2013-05-10 18:09:33

這對SSID的分區非常合適。非常感謝！ – m0ngr31 2013-05-10 18:26:49

獲取計數的總和與條件

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