考虑到下表,当profession.profession_id存在于contractor_has_profession.profession_id中时,我该如何编写返回profession.profession的查询,但将其限制为每个profession.profession只能返回一个结果
因此,在此示例中,结果将是Coder、Database、Frontend
contractor_has_profession
contractor_id | profession_id
1 | 5
2 | 5
3 | 5
4 | 2
5 | 1
profession
profession_id | profession
1 | Frontend
2 | Database
3 | Graphics
4 | Sound
5 | Coder发布于 2011-02-11 21:09:09
SELECT p.profession
FROM profession p
WHERE EXISTS(SELECT *
FROM contractor_has_profession c
WHERE c.profession_id = p.profession_id) 发布于 2011-02-11 21:12:06
嗯,这应该足够了:
select distinct p.profession
from profession p
inner join contractor_has_profession c
where p.profession_id = c.profession_id或者如果我说错了,那就试试:
select p.profession
from profession p
inner join contractor_has_profession c
where p.profession_id = c.profession_id
group by p.professionhttps://stackoverflow.com/questions/4969076
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