记录一下无聊的数据库作业

题目如下:

查询sC表中的全部数据。
2. 查询计算机系学生的姓名和年龄
3.查询成绩在70~80分的学生的学号、课程号和成绩
4.查询计算机系年龄在18~20岁的男生姓名和年龄
s.查询C001课程的最高分
6.查询计算机系学生的最大年龄和最小年龄
7.统计每个系的学生人数
8.统计每]课程的选课人数和最高成绩。
9.统计每个学生的选课门数和考试总成绩，并按选课]数升序显示结果。
10.列出总成绩超过200的学生的学号和总成绩
11.查询选了C002课程的学生姓名和所在系
12.查询考试成绩80分以上的学生姓名、课程号和成绩，并按成绩降序排列结果
13.查询与VB在同一学期开设的课程的课程名和开课学期
14.查询与李勇年龄相同的学生的姓名、所在系和年龄
15.查询哪些课程没有学生选修，列出课程号和课程名
16.查询每个学生的选课情况,包括未选课的学生，列出学生的学号、姓名、选的课程号
17.查询计算机系哪些学生没有选课，列出学生姓名
18.查询计算机系年龄最大的三个学生的姓名和年龄
19.列出“VB"课程考试成绩前三名的学生的学号、姓名、所在系和VB成绩
20.查询选课门]数最多的前2位学生，列出学号和选课门数

代码如下:

-- 1
SELECT *
FROM SC;
-- 2
SELECT s.Sname, s.Sage
FROM Student s
WHERE s.Sdept = N"计算机系";
-- 3
SELECT sc.Sno, sc.Cno, sc.Grade
FROM SC sc
WHERE sc.Grade BETWEEN 70 and 80;
-- 4
SELECT s.Sname, s.Sage
FROM Student s
WHERE s.Sdept = N"计算机系"
  AND s.Sage in (18, 20)
  AND s.Ssex = N"男";
-- 5
SELECT MAX(sc.Grade) AS max_grade
FROM SC sc
GROUP BY sc.Cno
HAVING sc.Cno = "C001";
-- 6
SELECT MAX(s.Sage) AS max_age, MIN(s.Sage) AS min_age
FROM Student s
GROUP BY s.Sdept
having s.Sdept = "计算机系";
-- 7
SELECT CONCAT(s.Sdept, " : ", COUNT(s.Sno)) AS stu_nums
FROM Student s
GROUP BY s.Sdept;
-- 8
SELECT sc.Cno AS Cno, COUNT(sc.Sno) as c_nums, MAX(sc.Grade) as max_grade
FROM SC sc
GROUP BY sc.Cno;
-- 9
SELECT COUNT(sc.Cno) as c_nums, SUM(sc.Grade) as sum_grades
FROM SC sc
GROUP BY sc.Sno
ORDER BY c_nums;
-- 10
SELECT sc.Sno, SUM(sc.Grade) AS sum_grades
FROM SC sc
GROUP BY sc.Sno
Having SUM(sc.Grade) > 200;
-- 11
SELECT s.sname, s.Sdept
FROM SC sc
         inner join Student s
                    on sc.Cno = "C002";
-- 12
SELECT s.Sname, sc.Cno, sc.Grade
FROM SC sc
         INNER JOIN Student s on sc.Sno = s.Sno
GROUP BY s.Sname, sc.Cno, sc.Grade
HAVING sc.Grade > 80
ORDER BY sc.Grade DESC;
-- 13
SELECT c.Cno, c.Semester
FROM Course c
WHERE c.Semester = (SELECT Semester FROM Course WHERE Cname = "VB")
  AND c.Cname <> "VB";
-- 14
SELECT s.Sname, s.Sdept, s.Sage
FROM Student s
WHERE s.Sage = (SELECT Sage FROM Student WHERE Sname = N"李勇")
  AND s.Sname <> N"李勇";
-- 15
SELECT c.Cno, c.Cname
FROM Course c
WHERE c.Cno not in (SELECT sc.Cno FROM SC sc);
--16
SELECT s.Sno,
       s.Sname,
       cno=STUFF((
                     SELECT "," + TRIM(c.Cno)
                     FROM Course c,
                          SC sc1
                     WHERE s.Sno = sc1.Sno
                       AND sc1.Cno = c.Cno
                     FOR XML PATH ("")), 1, 1, "")
FROM SC sc
         RIGHT JOIN Student S on sc.Sno = S.Sno
GROUP BY s.Sno, s.Sname;
-- 17
SELECT s.Sname
FROM Student s
WHERE s.Sno not in (SELECT sc.Sno FROM SC sc);
-- 18
SELECT
TOP 3
s.sname
,
s.Sage
FROM Student s
WHERE s.Sdept = N"计算机系"
ORDER BY s.Sage;
-- 19
SELECT
TOP 3
s.sno
,
s.sname
,
s.Sdept
,
sc.Grade
FROM Course c
         INNER JOIN SC sc ON c.Cno = sc.Cno
         INNER JOIN Student s on sc.Sno = s.Sno
WHERE c.Cname = "VB";
--20
SELECT
TOP 2
sc.Sno
,
COUNT(sc.Cno) AS course_nums
FROM SC sc
GROUP BY sc.Sno;