mysql - 尝试获取具有相应名称的最小值和最大值

Question

尝试在mysql中写入，以显示具有最大和最短时间的约会的Appointment_duration、mechanic_Firstname、mechanic_lastname、customer_firstname和customer_lastname，但我的代码给出了错误的结果。我只想要最大和最短持续时间以及相应的仅名称。所以 2 行

SELECT 
Min(Appointment.Appointment_duration) AS MINOfAppointment_duration, 
Max(Appointment.Appointment_duration) AS MAXOfAppointment_duration,
mechanic_Firstname, mechanic_lastname, customer_firstname, customer_lastname
FROM Appointment, mechanic, Customer
WHERE (mechanic.mechanic_ID=Appointment.mechanic_ID) AND (customer.customer_ID=Appointment.customer_ID);

预约表中的记录

Appointment_ID  Appointment_DATE    Appointment_Duration    Mechanic_ID Customer_ID
12               08/01/2007     0:35:00                      1            5684
13               01/01/2009     2:15:36                      6            2534
14               06/12/2010     0:05:29                      7            7423

Answer 1

问题是，两个表 customer 和 mechanic 之间没有关系，您必须分别获取每个表的最大和最短持续时间并使用 UNION ALL 将两个结果集合并为一个。类似于:

SELECT 
  m.mechanic_Firstname AS FirstName, 
  m.mechanic_lastname AS LastName,
  IFNULL(Min(a.Appointment_duration), 0) AS MINOfAppointment_duration, 
  IFNULL(Max(a.Appointment_duration), 0) AS MAXOfAppointment_duration
FROM mechanic AS m
LEFT JOIN Appointment AS a ON a.mechanic_ID = m.mechanic_ID
GROUP BY m.mechanic_Firstname,
         m.mechanic_lastname
UNION ALL
SELECT 
  c.customer_firstname,
  c.customer_lastname,
  IFNULL(Min(a.Appointment_duration), 0), 
  IFNULL(Max(a.Appointment_duration), 0)
FROM customer AS c
LEFT JOIN Appointment AS a ON a.mechanic_ID = c.customer_ID
GROUP BY c.customer_firstname,
         c.customer_lastname;

这只会给你四列:

FirstName  |  LastName  |  MINOfAppointment_duration  |  MAXOfAppointment_duration

如果所有机械师的姓名和客户的姓名都列在 firstname 和 lastname 两列中，您可以添加一个标记来标记来自客户的机械师。