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 |
---------------------------------------------------------------------

和第二桌学生

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

和第3桌讲座

-------------------------------------------------------------
| 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）进行查询并获取此tag_id并从学生表中获取有关它的完整信息。从students_absence中取出（讲座名= 1）并在讲课时将其与（id = 1）进行比较，并从中获取完整信息取决于id

最终结果应该是这样的

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

我无法得到一个查询给我上面的表，任何人都可以提供建议吗？

Answer 1

（未经测试）

select s1.tag_id s2.name,s3.lecture_name,s1.abs_hours from students_absence s1 join second_table s2 on s2.tag_id="00000023" and s1.tag_id="00000023" join lecture_table s3 on s1.lecture_name =s3.lecture_name

Answer 2

我假设在第一个表中你的意思是讲话而不是讲座名。如果我理解正确，那么输入就是并且您想知道某个班级学生失踪了多少小时。然后你需要的查询是这样的：

SELECT students.tag_id, name, lecture_name, lecturer_name, abs_hours FROM students_absence JOIN students on students_absence.tag_id = students.tag_id JOIN lecture_table on students_absence.lecture_id = lecture_table.id WHERE students.tag_id=<STUDENT_ID> AND lecture_table.id = <LECTURE_ID>

如果您需要所有班级中所有学生的列表，请省略＆＃34; WHERE＆＃34;部分和之后的一切

Answer 3

试试：

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

Answer 4

SQL命令：

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

Answer 5

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;

PHP / MySQL从3个不同的表中选择数据并将其显示在一个表中

5 个答案: