SQL查询以返回准确的子查询列表。

Question

所以我被问到这个问题：

考虑使用粗体显示键的下列表: Professor(profid，profname、department)、Student(studid，研究名称、专业)和Advise(profid，studid)。

返回与id为“123456789”的学生完全相同的顾问的学生的姓名。

我想出的问题不是返回完全相同的顾问，而是返回在学生123456789和其他学生之间常见的顾问。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。这个查询应该只返回同时有1和2两个顾问的学生。

SELECT studname
FROM Student
WHERE studid IN 
(
    SELECT DISTINCT studid
    FROM Advise
    WHERE profid IN
    (
        SELECT profid
        FROM Advise
        WHERE studid = '123456789'
    )
);

如何获得此查询以返回建议学生123456789的确切列表？

Answer 1

我测试它是否正确运行。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)

Answer 2

我的解决办法是找学生不一样的顾问，然后采取消极的这一点。

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')

Answer 3

你在找这个东西吗？

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' 
     )
  ) 
)