Question

所以我得到了这个问题：

考虑下面的表格，其中键是粗体：教授（ profid ，profname，department）学生（ studid ，studname，major）和Advise（ 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。

我提出的查询并未返回完全相同的顾问，而是返回学生123456789与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。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向学生123456789提供建议的学生的确切列表？

Answer 1

我测试它运行正确。你可以尝试：

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')

Answer 3

你在找这件事吗？

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' 
     )
  ) 
)

SQL查询返回子查询的确切列表

3 个答案: