Usage analysis of exists/not exists in SQL statements
1、Student：
Sno |
Take off |
Ssex |
Sage |
Sdept |
200215121 |
Li Yong |
male |
20 |
CS |
200215122 |
Liu Chen |
Female |
19 |
CS |
200215123 |
Wang Min |
Female |
18 |
MA |
200215124 |
Zhang Li |
male |
19 |
IS |
2、Course：
Cno |
Cname |
Cpno |
Credit |
1 |
database |
5 |
4 |
2 |
math |
|
2 |
3 |
Information system |
1 |
4 |
4 |
operating system |
6 |
3 |
5 |
data structure |
7 |
4 |
6 |
data processing |
|
2 |
7 |
PASCAL language |
6 |
4 |
3、SC：
Sno |
Cno |
Credit |
200215121 |
1 |
92 |
200215121 |
2 |
85 |
200215121 |
3 |
88 |
200215122 |
2 |
90 |
200215122 |
3 |
80 |
Example 1: Query the names of all students who have taken course No. 1.
Solution 1: Use exists
First take a tuple in the Student table, and then look for SC.Sno=Sno of the tuple in order in the SC table, and the corresponding Cno='1', if it exists, the where clause of the outer query returns true , the tuple in the Student table can be output. Then iterate over the other tuples in the Student table in turn.
For example: for the tuple whose number is equal to 2002151121 in the student table, the first record in the SC table is eligible, and then the where clause returns true, so the tuple can be output. Then traverse in turn.
select Sname
from Student
where exists
(
select *
from SC
where Sno = Student.Sno AND Cno='1'
);
Solution 2: Use join query
select Sname
from Student,SC
where Student.Sno=SC.Sno AND SC.Cno='1';
Example 2: Query the names of students who did not take course No. 1.
untie:
select Sname
from Student
where not exists
(
select *
from SC
where Sno=Student.Sno AND Cno='1'
);
Example 3: Query the names of students who have taken all courses.
select Sname
from Student
where not exists
(
select *
from Course
where not exists
(
select *
from SC
where Sno=Student.Sno AND
Cno=Course.Cno
) );
查找语义：查询这样的学生，没有一门课程是他不选修的。
查找过程：
首先，选取Student表中的一个元组，然后在依次判断Course表中的每个元组是否可以输出，只要有一个课程可以输出，则最外层查询的where子句返回为false；而在判断某个课程是否可以输出时，则要利用第三层查询，利用当前的学号和当前的课程号，在SC表中查询，如果存在，则第二层查询where子句返回false。至此，每一门课程都不可以输出时，这个学号对应的元组才可以输出。表示这个学生选修了全部的课程。
例4：至少选修了学生200215122选修的全部课程的学生号码。
select distinct Sno
from SC SCX
where not exists
(
select *
from SC SCY
where SCY.Sno='2002151122' AND
not exists
(
select *
from SC SCZ
where SCZ.Sno=SCX.Sno AND
SCZ.Cno=SCY.Cno));
查询语义：不存在这样的课程y，学生200215122选修了y，而学生x没选。
查询过程：先在SCX表中选一条记录，比方说第一条，然后再看SCY表中，只有SCY表中全部不能输出，第一层查询的where子句才返回true，第一条记录就可以输出；所以就要一次查看SCY表中的每一个元组，前三个，因为学号首先不满足=200215122所以必然不能输出，第四个的话，就要看其AND后面的not exists返回什么值，而这又取决于第三层查询中是否存在满足学号等于SCX.Sno且课程号=SCY.Cno的元组，经查看，有 ，则返回false，所以第四个也不能输出，第五个类似，所以，第一层查询的not exists返回true。所以第一条记录可以输出。