Then notes training 3
For me the introduction a more difficult difficulty ! ! ! !
select * from sc;
the SELECT student.sno, sname, AVG (Grade) AS average from Student, sc the WHERE student.sno = sc.sno Group by student.sno the HAVING AVG (Grade) > 90 / * This is a condition I actually forgot to write qwq * / the Order by Grade desc ;
I just forgot to write when the average score higher than 90 points this condition a ! ! ! ! ! ! ! ! ! ! !
After grouping there is a having to filter it! ! ! ! ! !
having clause and where there are similarities but also differences, all statements setting conditions.
having a screening group and where is screening record.
go on----------->
select student.sno, sname, avg (grade) as Average from student,sc where student.sno=sc.sno group by student.sno having avg (grade)> 90 / * This is a condition I actually forgot to write qwq * / order by grade desc limit 2;
2. Next start connecting from outside connections, fully connected like it, study and understand the master side of the SQL language of relational algebra language Oh, 2333!
Since the connection :
select * from student;
/ * Query the name and cloudy in the same Academy classmates * / select s2.sname from student as s1,student as s2 where s1.sname = 'cloudy' and s1.sdeptno=s2.sdeptno;
Indirect queries each course Prerequisite
SELECT c1.cno, c1.cname, c2.cpre AS cppre from C AS C1, C AS C2 WHERE c1.cpre = c2.cno; / * find each course indirect Elective * /
Says a relevant, da da oh:
Such a table may be provided an outer code
The curriculum is the outer code Prerequisite curriculum! ! ! ! ! ! !
https://www.cnblogs.com/xiohao/archive/2013/06/28/3160265.html
ALTER Table c the Add constraint FK_c Foreign Key (cpre) References c (CNO); / * the curriculum field is set to c, c cpre outer code, a reference relationship is c * /
The next question -------- "
Code and the results are as follows:
SELECT c1.cno, c1.cname, c2.cpre AS cppre from C AS C1, C AS C2 WHERE c1.cpre = c2.cno; / * find each course indirect Elective * /
/ * Left outer join * / / * query the case for all students ... * / the SELECT sname, CNO, Grade from Student left the Join sc ON student.sno = sc.sno;
Right outer join
/ * Query all course information, ... * / the SELECT CNAME, sc.sno, Grade from sc right Outer the Join c ON sc.cno = c.cno; / * the right connection * /
