Question

在我的mysql数据库中，我有以下表格：

FACULTY (fid int, fname varchar(25), deptid int, primary key(fid))  
CLASS (name varchar(4),meets_at varchar(9),room varchar(4), fid int,primary key (name), foreign key (fid) references faculty (fid))

我想选择去所有房间的院系名称。我尝试使用以下内容：

SELECT DISTINCT F.FNAME   
FROM FACULTY F  
WHERE NOT EXISTS (( SELECT *  
FROM CLASS C  
EXCEPT  
(SELECT C1.ROOM  
FROM CLASS C1  
   WHERE C1.FID=F.FID)));

并收到以下错误：

错误1064（42000）：您的SQL语法有错误;检查手册，对应右边的MySQL服务器版本在'EXCEPT
附近使用的语法

也尝试过：

SELECT DISTINCT F.FNAME   
FROM FACULTY F   
LEFT JOIN CLASS C ON C.FID = F.FID  
WHERE C.FID IS NULL

即使在我的数据库中有一个教职员去所有房间，也得到了“空集”。

Answer 1

当您使用except时，两个表格必须为compatible，请尝试以下操作：

SELECT DISTINCT F.FNAME   
FROM FACULTY F  
WHERE NOT EXISTS (
( SELECT ROOM  
FROM CLASS C)
EXCEPT  
(SELECT C1.ROOM  
FROM CLASS C1  
   WHERE C1.FID=F.FID)
);

修改

问题被标记为sql server所以我给出了答案，记住，为mysql使用这个：

SELECT FID, COUNT(*) FROM ( (SELECT DISTINCT f.fname, f.fid, c1.room FROM faculty f JOIN class c1 ON f.fid = c1.fid) tb1 JOIN (SELECT DISTINCT room AS room2 FROM class) tb2 ON tb1.room = tb2.room2 ) GROUP BY FID HAVING COUNT(*) IN (SELECT COUNT(DISTINCT Room) FROM Class);

小提琴：http://sqlfiddle.com/#!8/cff12/4

Answer 2

我的观点与你的问题有关。

SELECT T.fid FROM
  (SELECT f.fid FROM 
     faculty f LEFT JOIN class c ON f.fid = c.fid) T
  GROUP BY T.fid 
  HAVING(
    COUNT(T.fid) = (SELECT COUNT(DISTINCT room) FROM class))

SQL fiddle

从列的表中选择所有内容

2 个答案: