我正在为学校设计一个基于网络的时钟,用于登录学生和员工。
我有两张桌子:
student_time_clock: student_id (foreign key to student), time_stamp, by_adult_id (foreign key to adult)
staff_time_clock: staff_id (foreign key to staff), time_stamp
我给出的一个要求是观察者能够在一个视图中看到如下记录:
+----------+------------+-------------------------+---------------------+
| In / Out | Time Stamp | Person Logged in or out | Logged in or out by |
+----------+------------+-------------------------+---------------------+
除了带有联合的第一列,我可以做所有事情。我无法弄清楚如何获得第一列。这是我用于联合的查询:
SELECT stc.entry AS "Time Stamp",
Concat(s.lastname, ",", s.firstname) AS "Punched",
Concat(a.lastname, ",", a.firstname) AS "By"
FROM student_time_clock stc,
student s,
adult a
WHERE stc.student_id = s.id
AND stc.by_adult_id = a.id
UNION
SELECT atc.entry AS "at",
Concat(a.lastname, ",", a.firstname) AS "Staff",
Concat(a.lastname, ",", a.firstname) AS "By"
FROM staff_time_clock atc,
staff s,
adult a
WHERE atc.staff_id = s.id
ORDER BY "time stamp" DESC;
我尝试使用CASE,例如:
CASE COUNT( entry ) % 2 WHEN 1 THEN "In" WHEN 0 THEN "Out"
当那种情况出现时,我只在结果中得到一行。
有什么建议吗?前端是PHP,但我想将其创建为数据库中的视图
答案 0 :(得分:0)
我认为最好在第一个表中添加一个标志列来捕获类型或条目,即“In”或“Out”。这可以通过某些结果简化您的查询。