我有3张桌子
现在我想得到
GroupID,UserName,MAX(PaymetDate),MAX(AttendenceDate)
MAX(PaymetDate)小于MAX(AttendenceDate)
SELECT MAX(PaymetDate) AS Paied_Upto
FROM Payment
Group by GroupID
SELECT MAX(AttendenceDate) AS Last_ AttendenceDate
FROM Attendence FULL OUTER JOIN Users ON Attendence.Username = Users.Username
Group by Users.GroupID
但是如何让他们一起工作?
答案 0 :(得分:1)
试试这个:
SELECT u.GroupID, u.UserName, py.LastPaymentDate, at.LastAttendenceDate
FROM User AS u,
(SELECT Username, Max(AttendenceDate) AS LastAttendenceDate FROM Attendence GROUP BY Username) AS at,
(SELECT GroupID, Max(PaymetDate) AS LastPaymentDate FROM Payment GROUP BY GroupID) AS py
WHERE u.UserName=at.Username
AND u.GroupID=py.GroupID
AND py.LastPaymentDate < at.LastAttendenceDate;
答案 1 :(得分:0)
试试这个
select p.GroupID, u.UserName, MAX(p.PaymetDate), MAX(a.AttendenceDate)
from dbo.Users u
inner join dbo.Attandence a
ON u.UserName = a.UserName
Inner join dbo.Payment p
ON u.groupID = p.GroupID
GROUP BY p.GroupID, u.UserName
Having MAX(p.PaymentDate) < MAX(a.attendenceDate)
答案 2 :(得分:0)
我认为这可以满足您的需求( SqlFiddle link ):
select UserName, GroupID, MaxAttendanceDate, MaxPaymentDate
from (
select
u.UserName,
u.GroupID,
(select max(AttendanceDate)
from Attendance a
where a.UserName = u.UserName) as MaxAttendanceDate,
(select max(PaymentDate)
from Payment p
where p.GroupID = u.GroupId) as MaxPaymentDate
from [User] u
) x
where MaxAttendanceDate > MaxPaymentDate