2017-08-10 78 views
1

所以我在這個問題上掙扎了一下,我在基於小時的學校做入門級工作。SQL,基於變量的日期X和X之間的總和

我需要找到所有已通過註冊結束日期的學生,並從他們的第一次出勤日期到註冊結束日期獲得這些學生小時的總和。這是我寫的代碼的開始,但我知道我的Case是錯誤的。

示例學生A的出勤日期大於其合同結束日期。

在合同結束日期他的總小時數= 560

最小完成此計劃爲600小時

然後,我們添加一個附加費,以這40個小時,他已經沒有結束日期完成。

我可以做數學,我只需要得到它,我可以在那裏讓學生身體。如果我添加諸如ATD.Attendancedate> = Reg.enddate之類的任何內容,它當然只會在結束日期之前拉動幾小時。

SELECT CONCAT('<a href="admin_view_student.jsp?studentid=', 
    CAST(SDT.studentId AS CHAR), '">', CAST(SDT.firstName AS bCHAR), ' ', 
    CAST(SDT.lastName AS CHAR), '</a>') AS Name, 
    PGM.programmeName AS 'Program', 
    FORMAT(SUM(ATD.duration),2) AS 'Total Hours', 
    PGM.MinClockHours AS 'Program Total', 
    ATD.attendancedate AS 'ATTDATE', 
REG.Enddate AS 'EndDate', 
Case WHEN MAX(ATD.attendancedate) >= REG.Enddate Then Sum(ATD.duration) Between MIN(ATD.attendancedate) AND REG.enddate 
ELSE NULL END AS 'Whatever' 
FROM Attendance ATD 
INNER JOIN Registrations REG ON ATD.studentId = 
     REG.studentId AND REG.isActive = 1 
INNER JOIN Programmes  PGM ON REG.programmeId 
    = PGM.programmeId AND PGM.isActive = 1 
INNER JOIN Students   SDT ON REG.studentId = 
    SDT.studentId AND SDT.isactive = 1 
    WHERE 
    REG.ADMINID       AND 
    REG.enrollmentSemesterId = 4000441 AND 
    PGM.programmename Not like ('Careers Pathway') AND PGM.programmename Not like ('Instructor Training') AND 
    ATD.subjectId IN (SELECT GSR.subjectId 
      FROM CourseGroups CGP 
      INNER JOIN GroupSubjectReltn GSR ON CGP.courseGroupId=GSR.courseGroupId AND GSR.isActive=1 
      WHERE REG.programmeId = CGP.programmeId and CGP.isActive=1) 
             AND 
    ATD.classId IN (SELECT DISTINCT CRS.classId 
      From ClassStudentReltn CRS 
      Where CRS.studentId = ATD.studentId AND CRS.isActive=1) 
GROUP BY SDT.lastname 
ORDER BY SDT.firstName 
+0

*不確定包含SUM()的情況表達式通常它的工作方式是相反的。例如* SUM(Case WHEN MAX(ATD.attendancedate)> = REG.Enddate and ATD.attendancedate MIN(ATD.attendancedate)AND REG.enddate then ATD.duration end) –

回答

0

您可以通過自聯接方式直接解決這個問題。您需要根據Where ATD.Attendancedate >= Reg.enddate選擇有問題的學生,然後僅對滿足該條件的學生進行計算。所以:

[your query] 
inner join 
    (SELECT studentId 
    FROM Attendance ATD 
    inner join Registrations REG 
    ON ATD.studentId = REG.studentId 
     AND REG.isActive = 1 
    WHERE ATD.Attendancedate >= Reg.enddate) as late 
on [your query].studentId = late.studentId