1、Student(S#,Sname,Sage,Ssex) 学生表S#：学号； Sname：学生姓名；Sage ：学生年龄；Ssex：学生性别Course(C#,Cname,T#) 课程表C#,课程编号；Cname：课程名字； T#：教师编号SC(S#,C#,score) 成绩表S#：学号； C#,课程编号；score ：成绩Teacher(T#,Tname) 教师表 T#：教师编号； Tname：教师名字13、查询学生平均成绩及其名次(不用Row_num()1、查询“001”课程比“002” 课程成绩高的所有学生的学号；1、查询“001”课程比“002” 课程成绩高的所有学生的学号；SELECT
2、 a.s# FROM ( SELECT s# ,score FROM sc WHERE c#=001) a ,( SELECT s# ,score FROM sc WHERE c#=002) b WHERE a.scoreb.score AND a.s#=b.s#;2、查询平均成绩大于60分的同学的学号和平均成绩；2、查询平均成绩大于60分的同学的学号和平均成绩；SELECT s# ,avg ( score ) FROM sc GROUP BY s# having avg ( score ) 60;3、查询所有同学的学号、姓名、选课数、总成绩；3、查询所有同学的学号、姓名、选课数、总成绩；SE
3、LECT student.s# ,student.sname ,COUNT ( sc.c# ) ,SUM ( score ) FROM student LEFT OUTER JOIN sc ON student.s#=sc.s# GROUP BY student.s# ,sname4、查询姓“李”的老师的个数；4、查询姓“李”的老师的个数；SELECT COUNT ( DISTINCT ( tname ) ) ，COUNT ( tname ) FROM teacher WHERE tname LIKE 李%;SELECT COUNT ( tname ) FROM teacher WHERE t
4、name LIKE 李%;5、查询没学过“ 叶平”老师课的同学的学号、姓名；5、查询没学过“ 叶平”老师课的同学的学号、姓名；SELECT student.s# ,student.sname FROM student WHERE s# NOT IN ( SELECT DISTINCT ( sc.s# ) FROM sc ,course ,teacher WHERE sc.c#=course.c# AND teacher.t#=course.t# AND teacher.tname=叶平 ) ;6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；6、查询学过“001”并且也学过
5、编号“002”课程的同学的学号、姓名；SELECT student.s# ,student.sname FROM student ,sc WHERE student.s#=sc.s# AND sc.c#=001 and EXISTS ( SELECT * FROM sc AS sc_2 WHERE sc_2.s#=sc.s# AND sc_2.c#=002 ) ;7、查询学过“叶平”老师所教的所有课的同学的学号、姓名；7、查询学过“叶平”老师所教的所有课的同学的学号、姓名；SELECT s# ,sname FROM student WHERE s# IN ( SELECT s# FROM sc
6、 ,course ,teacher WHERE sc.c#=course.c# AND teacher.t#=course.t# AND teacher.tname=叶平 GROUP BY s# having COUNT ( sc.c# ) = ( SELECT COUNT ( c# ) FROM course ,teacher WHERE teacher.t#=course.t# AND tname=叶平 ) ) ;8、查询所有课程成绩小于60 分的同学的学号、姓名；8、查询所有课程成绩小于60 分的同学的学号、姓名；SELECT s# ,sname FROM student WHERE s
7、# NOT IN ( SELECT student.s# FROM student ,sc WHERE s.s#=sc.s# AND score60 ) ;9、查询没有学全所有课的同学的学号、姓名；9、查询没有学全所有课的同学的学号、姓名；SELECT student.s# ,student.sname FROM student ,sc WHERE student.s#=sc.s# GROUP BY student.s# ,student.sname having COUNT ( c# ) t2.平均成绩 ) 名次 ,s# 学生学号 ,平均成绩 FROM ( SELECT s# ,avg (
8、score ) 平均成绩 FROM sc GROUP BY s# ) t2 ORDER BY 平均成绩 desc;（这里用了 COUNT ( DISTINCT 平均成绩 ) ，考虑了并列相同名次的情况）14、查询各科成绩前三名的xuehao记录:( 不考虑成绩并列情况 )14、查询各科成绩前三名的记录:( 不考虑成绩并列情况)SELECT t1.s# AS 学生id ,t1.c# AS 课程id ,score AS 分数 FROM sc t1 WHERE score IN ( SELECT top 3 score FROM sc WHERE t1.c#= c# ORDER BY score DESC ) ORDER BY t1.c#;15、查询每门功成绩最好的前两名15、查询每门功成绩最好的前两名SELECT t1.s# AS 学生id ,t1.c# AS 课程id ,score AS 分数 FROM sc t1 WHERE score IN ( SELECT top 2 score FROM sc WHERE t1.c#= c# ORDER BY score DESC ) ORDER BY t1.c#;(如果成绩有相同的情况，就可能出现多余两名的情况)
