PHP/MySQL 从 3 个不同的表中选择数据并将结果显示在一张表中

Question

我的数据库中有 3 个表

这是students_absence表

---------------------------------------------------------------------
| tag_id    | lecture_name  | abs_hours | abs_ratio     | days      |
|-----------+---------------+-----------+---------------+-----------|
| 00000023  | 1             | 8         | 5.3333        | 13/4/2014 |
---------------------------------------------------------------------

还有二 table 的同学

---------------------------------------------------------------------------------
| tag_id        | name      | collage       | department    | class | gender    |                 |
|---------------+-----------+---------------+---------------+-------+-----------+
| 00000023      | John Deem | Engineering   | Computer      | first | male      |  
---------------------------------------------------------------------------------

和第三个表lecture_table

-------------------------------------------------------------
| id    | lecture_name  | lecturer_name     | start | end   |
|-------+---------------+-------------------+-------+-------+
| 1     | Math 1        | Dr. Anas Ali      | 8     | 10    |
-------------------------------------------------------------

所以我需要在students_absence表中通过tag_id(00000023)进行查询并获取这个tag_id并从students表中获取有关它的完整信息。并从students_absence中获取(lecture_name = 1)并将其与lecture_table中的(id = 1)进行比较，并根据id从中获取完整信息

最终的结果应该是这样的

---------------------------------------------------------------------------------
| tag_id        | name      | lecture_name  | lecturer_name     | abs_hours     |       
|---------------+-----------+---------------+-------------------+---------------+
| 00000023      | John Deem | Math 1        | Dr. Anas Ali      | 8             |
---------------------------------------------------------------------------------

我无法查询，请给我上表，任何人都可以提供建议吗？

Answer 1

您需要使用join

SELECT s.tag_id, s.name, lt.lecture_name, lt.lecturer_name, sa.abs_hours
FROM students_absence sa
JOIN students s ON (s.tag_id = sa.tag_id)
JOIN lecture_table lt ON (s.lecturer_name = lt.id)
WHERE s.tag_id = 00000023;