Question

这个令我困惑？

有三张桌子。 Appointments(Appointment_ID, Physician_ID and Person_ID)，Physician(Physician_ID)和Person(Person_ID and Physician_ID)。

这是我到目前为止所做的：

SELECT DISTINCT Appointment_date_time FROM Appointment 
INNER JOIN Person 
        ON Appointment.Person_ID = Person.Person_ID
INNER JOIN Physician 
        ON Physician.Physician_ID = Person.Physician_ID
HAVING COUNT(*) < 1

Answer 1

有三张桌子。约会（Appointment_ID，Physician_ID和Person_ID），Physician（Physician_ID）和Person（Person_ID和Physician_ID）。

select *
from Appointments a
inner join Person p
on a.Person_ID = p.Person_ID
inner join Physician ph
on a.Physician_ID = ph.Physician_ID

编写将仅列出ONCE的Physician-Person约会的SQL代码

1 个答案: