Question

我有两张表，比如

表1：FACULTY_DETAILS

fac_det_id （pk）........ fname ........... 可用性 < / p>

..... 1 ......................... xxx ............... ..完整的时间

..... 2 ......................... yyy ............... ..部分时间

..... 3 ......................... zzz ............... ..weekdays

..... 4 ......................... aaa ............... ..partime

表2：FACULTY

faculty_id （PK）..........的 COURSE_ID ........的 fac_det_id （FK ）

..... 1 .............................. 1 .......... ............ 2

..... 2 .............................. 2 .......... ............ 3

..... 3 .............................. 3 .......... ............ 1

..... 4 .............................. 4 .......... ............ 3

..... 5 .............................. 3 .......... ............ 4

当我给出课程ID时我需要fname，table1中的可用性和表2中的faculty_id

即如果我给出course_id = 3那么我需要

faculty_id ..........的 FNAME ...........的资格

..... 2 ................... xxx ................全职

..... 5 ................... AAA ................ partime

Answer 1

试试这个：

SELECT f.faculty_id, fd.fname, fd.availability  qualification 
FROM FACULTY f 
INNER JOIN FACULTY_DETAILS fd ON f.fac_det_id = fd.fac_det_id 
WHERE f.course_id = 3

Answer 2

您可以使用：

SELECT faculty_id, fname, availibility AS qualification
FROM faculty a, faculty_details b
WHERE a.fac_det_id = b.fac_det_id AND course_id = $your_course_id;

Answer 3

这应该有效

select f.faculty_id,fd.fname,fd.availability as qualification
from faculty f,faculty_details fd
inner join  faculty_details fd ON f.fac_det_id = fd.fac_det_id 
where course_id=3;

mysql的复杂选择查询将两列与where子句组合在一起

3 个答案: