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我正在为一所学校设计一个基于网络的时钟，用于登录学生和教职员工。

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

我得到的要求之一是观察者能够在一个视图中查看记录，如下所示：

+----------+------------+-------------------------+---------------------+
| In / Out | Time Stamp | Person Logged in or out | Logged in or out by |
+----------+------------+-------------------------+---------------------+

除了第一列之外，我可以用工会做所有事情。我不知道如何获得第一列。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;

我尝试过使用 CASE，例如：

CASE COUNT( entry ) % 2 WHEN 1 THEN "In" WHEN 0 THEN "Out"

当这种情况存在时，尽管我在结果中只得到一行。

有什么建议么？前端是 PHP，但我想在数据库中创建它作为视图

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1 回答 1

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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。

于 2013-05-24T15:39:55.637 回答