[英]SQL query to return an exact list of subquery
所以我得到了这个问题:
考虑以下表键进行加粗:教授(PROFID,profname,系)学生(studid,studname,主要),并建议(PROFID,studid)。
返回与id
为“123456789”的学生具有完全相同顾问的学生的姓名。
我提出的查询并没有返回完全相同的顾问,而是返回学生123456789和其他学生之间共同的顾问。 例如,如果学生123456789有顾问1和2,而学生5只有顾问1,我当前的查询将返回学生5,这是不正确的。 该查询只能返回同时拥有顾问1和2的学生。这是我目前的查询:
SELECT studname
FROM Student
WHERE studid IN
(
SELECT DISTINCT studid
FROM Advise
WHERE profid IN
(
SELECT profid
FROM Advise
WHERE studid = '123456789'
)
);
如何才能获得此查询以返回建议学生123456789的学生的确切列表?
我测试它运行正确。 你可以试试:
SELECT a.studid, b.studname
FROM (
SELECT studid, COUNT(studid) AS numstud
FROM Advise
WHERE
profid IN (
SELECT profid FROM Advise WHERE studid = 123456789
) AND
studid NOT IN (
SELECT studid FROM Advise WHERE profid NOT IN (
SELECT profid FROM Advise WHERE studid = 123456789
)
)
GROUP BY studid
HAVING numstud = (SELECT COUNT(*) FROM Advise WHERE studid = 123456789)
) AS a LEFT JOIN Student AS b ON (a.studid = b.studid)
我的解决方案是找到学生不同的顾问,然后对此不予理睬。
请尝试使用脚本:
SELECT studid,studname
FROM Student
WHERE studid not in
(
select studid
from
(
SELECT a.studid, b.profid
FROM Advise a
left join
(
SELECT profid
FROM Advise
WHERE studid = '123456789'
) b on b.profid = a.profid
where a.studid not like '123456789'
) x
where x.profid is null
)
group by studid,studname
having count(*) = (SELECT count(profid) FROM Advise WHERE studid = '123456789')
你在寻找一些东西吗?
select studname from Student where studid in
(
select studid from (
select studid, GROUP_CONCAT(profid) profs from advice
group by studid having profs in (
select studid, GROUP_CONCAT(profid) profs from advice group by
studid having studid = '123456789'
)
)
)
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