大家好,我是考100分的小小码 ,祝大家学习进步,加薪顺利呀。今天说一说数据库作业在哪里存着_mysql数据库大作业,希望您对编程的造诣更进一步.
题目如下:
1.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;
代码100分
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