[MySQL练习] 查询两个表的差集（两种方法）

在经典sql50题中这道题是这样的：

有三张表

student表

 course表

 score表

#10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息

这是网上查的

select st.* from student st 
inner join score sc on sc.s_id = st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="01"
where st.s_id not in (
select st2.s_id from student st2 
inner join score sc2 on sc2.s_id = st2.s_id
inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02"
)

这是我自己写的（单纯觉得这样看上去清楚一点，然后就懒得跟课程表连了，有成绩的话他就学了，没有成绩他就没学）

第一种方法就是用IN然后用学生表中s_id跟成绩表中条件是02课程的s_id 进行比较

SELECT t1.*
FROM student t1  
JOIN (SELECT s_id    
      FROM score 
      WHERE c_id =01 and s_id NOT in (SELECT s_id                                                    
                                      FROM score                                                                 
                                      WHERE c_id =02)) t2 
ON t1.s_id = t2.s_id                                      

第二种方法就是

用学习01课程的表左连接学习02课程的表查询出来s_id，因为是左连接学过01但没有学过02的人他的那个成绩就是用NULL显示

根据这个NULL进行判断。如果是NULL他就是学过01但没学过02的。

SELECT t1.*
FROM student t1 JOIN 
(SELECT t1.s_id
FROM (SELECT s_id,s_score    
FROM score 
WHERE c_id =01) t1 
LEFT JOIN 
(SELECT s_id,s_score                                                                
FROM score 
WHERE c_id =02) t2
ON t1.s_id = t2.s_id 
WHERE t2.s_score IS NULL) t2 
ON t1.s_id = t2.s_id

查出来结果都是一样的

 记录一下哈哈