SQL基本查询大综合

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SQL

Student(S#,Sname,Sage,Ssex) 学生表
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 教师表

CREATE TABLE `test`.`Student` (
`S#` int(10) unsigned NOT NULL ,
`Sname` varchar(45) NOT NULL,
`Sage` int(10) NOT NULL,
`Ssex` varchar(45) NOT NULL,
PRIMARY KEY (`S#`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `test`.`Course` (
`C#` int(10) unsigned NOT NULL ,
`Cname` varchar(45) NOT NULL,
`T#` int(10) NOT NULL,

PRIMARY KEY (`C#`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `test`.`SC` (
`S#` int(10) unsigned NOT NULL ,
`C#` int(10) unsigned NOT NULL ,
`score` int(10) unsigned NOT NULL ,
PRIMARY KEY (`S#`,`C#`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8

CREATE TABLE `test`.`Teacher` (
`T#` int(10) unsigned NOT NULL ,
`Tname` varchar(45) NOT NULL,
PRIMARY KEY (`T#`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `Student` (`S#`,`Sname`,`Sage`,`Ssex`) VALUES
(1,'张三',15,'男'),
(2,'李四',16,'男'),
(3,'王五',15,'男'),
(4,'马六',16,'男'),
(5,'晨七',17,'男'),
(6,'丽丽',15,'女'),
(7,'婷婷',16,'女'),
(8,'娜娜',15,'女'),
(9,'欢欢',16,'女'),
(10,'泉灵',17,'女');

INSERT INTO `Student` (`T#`,`Tname`) VALUES
(1,'老师1'),
(2,'老师2');

INSERT INTO `Course` (`C#`,`Cname`,`T#`) VALUES
(1,'001',1),
(2,'002',2);

INSERT INTO `SC` (`S#`,`C#`,`score`) VALUES
(1,'001',90),
(2,'001',90),
(3,'001',80),
(4,'001',79),
(5,'001',85),
(6,'001',60),
(7,'001',59),
(8,'001',85),
(9,'001',95),
(10,'001',95),
(1,'002',90),
(2,'002',90),
(3,'002',80),
(4,'002',79),
(5,'002',85),
(6,'002',60),
(7,'002',59),
(8,'002',85),
(9,'002',95),
(10,'002',95);

问题：
1)、查询“001”课程比“002”课程成绩高的所有学生的学号；
2)、查询平均成绩大于60分的同学的学号和平均成绩；
3)、查询所有同学的学号、姓名、选课数、总成绩；
4)、查询姓“李”的老师的个数；
5)、查询没学过“叶平”老师课的同学的学号、姓名；
6)、查询学过“001”并且也学过编号“002”课程的同学的学号、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；
8)、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；
9)、查询所有课程成绩小于60分的同学的学号、姓名；
10)、查询没有学全所有课的同学的学号、姓名；

1) select a.S# from (select s#,score from SC where Cname='C601') a,(select s#,score
    from SC where Cname='C602') b
    where a.score>b.score and a.s#=b.s#;
    //--------------
    select a.s#
    from sc a
    where a.c#='C601' and exists(select * from sc b where b.score<a.score and b.c#='C602' and a.s#=b.s#)
补充1：
   select a.`S#` from
   (select `s#`,score from SC where `C#` in (select `C#` from Course where `Cname`='001') )a,
   (select `s#`,score from SC where `C#` in (select `C#` from Course where `Cname`='002') )b
    where a.score>b.score and a.`s#`=b.`s#`;
    //--explain :
        //--1)在SC表中查询所有课程名Cname=001（在课程表Course中确定）的课程主键S#的结果集 a
        //--2) 同1),查询所有课程名Cname=001的课程主键S#的结果集 b
        //--3)
    //--------

    select a.`s#`
    from sc a
    where a.`c#` in (select `C#` from Course where `Cname`='001')
    and exists(
    select *
    from sc b
    where b.score<a.score
    and b.`c#` in (select `C#` from Course where `Cname`='002')
    and a.`s#`=b.`s#`);


2) select S#,avg(score)
      from sc
      group by S#
      having avg(score) >60;
      补充：
      select sc.`s#` 学号 ,avg(sc.score) 平均分
       from sc
       group by sc.`s#`
       having avg(sc.score) >80


3) select Student.S#,Student.Sname,count(SC.C#),sum(score)
    from Student left Outer join SC on Student.S#=SC.S#
    group by Student.S#,Sname

   补充：
   select student.`s#` 学号,student.sname 姓名,count(sc.`c#`) 选课数,sum(sc.score) 总成绩
       from student left outer join sc
       on student.`s#` = sc.`s#`
       group by student.`s#`
4) select count(distinct(Tname))
    from Teacher
    where Tname like '李%';

   补充：
   select count(*)   姓李的老师数
       from teacher t
       where t.tname
       like '李%'


5) 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');

    select Student.S#,Student.Sname
    from Student
    where not exists(
   select *
   from SC,Course,Teacher
   where Student.S#=SC.S#
   and SC.C#=Course.C#
   and Teacher.T#=Course.T#
   and Teacher.Tname='叶平')


   补充：

   select student.`s#`,student.sname
   from student
   where student.`s#`
   not in (
   select student.`s#`
   from student,sc,course,teacher
   where student.`s#` = sc.`s#`
   and sc.`c#` = course.`c#`
   and course.`t#` = teacher.`t#`
   and teacher.tname = '张亮')


6) 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');

       补：
       select student.`S#` 学号 ,student.Sname 学生名
           from student,course,sc
           where student.`S#` = sc.`S#`
               and sc.`C#`= course.`C#`
               and course.Cname = '001'
               and student.`S#`
               in (select student.`S#`
                       from student,course,sc
                       where student.`S#` = sc.`S#`
                       and sc.`C#` = course.`C#`
                       and course.Cname = '002')


7) select S#,Sname
    from Student
    where S#
   in (
   select S#
   from SC ,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'));





    select S#,Sname
    from Student
    where not exists(select * from Course where exists(select * from Teacher where Teacher.Tname='叶平' and Course.T#=Teacher.T# and not exists(select * from SC where SC.C#=Course.C# and SC.S#=Student.S#)))
    --找出没有没选所有课的学生而且这个课是叶平老师教的，即找出有选叶平老师教的所有课的学生

   补：
   select student.`S#` 学号 ,student.Sname 学生名,count(course.Cname) 课程数
       from student ,course , sc ,teacher
       where student.`S#` = sc.`S#`
       and sc.`C#`= course.`C#`
       and course.`T#` = teacher.`T#`
       and teacher.Tname = '李红'
       group by student.`S#`
       having count(course.Cname) = (
                                       select count(course.Cname)
                                       from course,teacher
                                       where course.`T#` = teacher.`T#`
                                       and teacher.Tname = '李红')

8) select a.S# from (select s#,score from SC where C#='001') a,(select s#,score
    from SC where C#='002') b
    where a.score>b.score and a.s#=b.s#;

9) select Student.S#,Student.Sname
    from Student
    where Student.S# not in (select SC.S# from SC where Student.S#=SC.S# and SC.score>60)



    select Student.S#,Student.Sname
    from Student
    where not exists(select * from SC where Student.S#=SC.S# and SC.score>60)
10)select Student.S#,Student.Sname
    from Student,SC
    where Student.S#=SC.S#
   group by Student.S#,Student.Sname
   having count(C#) <(select count(C#) from Course);

    select S#,Sname
    from Student
    where exists(select * from Course where not exists (select * from SC where SC.S#=Student.S# and SC.C#=Course.C# ))
    --找出存在没有选所有课的学生，即：找出没有学选所有课的学生


   补：
   select student.`S#` 学号,student.Sname 学生名,count(sc.`C#`) 课程数
    from sc , student
   where sc.`S#` = student.`S#`
   group by sc.`S#`
   having count(sc.`C#`)<> (select count(course.`C#`)
                           from course)

11)、查询至少有一门课与学号为“98602”的同学所学相同的同学的学号和姓名；
新补：查询至少有一门课比学号为“98602”的同学所学相同的分高同学的学号和姓名；
12)、查询至少学过学号为“98602”同学所有一门课的其他同学学号和姓名；
13)、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩；(有问题)
14)、查询和“98606”号的同学学习的课程完全相同的其他同学学号和姓名；(有问题)
15)、删除学习“叶平”老师课的SC表记录；
16)、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“002”课程的同学学号、002号课程的平均成绩；
17)、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩，按如下形式显示：学生ID,,数据库,企业管理,英语,有效课程数,有效平均分
18)、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分
19)、按各科平均成绩从低到高和及格率的百分数从高到低顺序
20)、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理（001），自然辩证法（002），OO&UML （003），数据库（004）

11) select distinct(Student.S#),Student.Sname from Student,SC where Student.S#=SC.S# and SC.C# in (select C# from SC where S#='98603');

       select distinct(Student.S#),Student.Sname from Student,SC where Student.S#=SC.S# and exists (select * from SC b where SC.C#=b.C# and b.S#='98603');
       select Student.S#,Student.Sname from Student where exists(select * from SC where Student.S#=SC.S# and exists (select * from SC b where SC.C#=b.C# and b.S#='98603'));

   --找出存在选过学号为98603同学选的课的学生

   补：
   select distinct student.`S#`,student.Sname
   from student,sc
   where student.`S#` = sc.`S#`
   and student.`S#`
   in(
       select sc.`C#`
       from sc
       where sc.`S#` = 1
   )



   新补：
   select distinct student.`S#`,student.Sname
   from student,sc
   where student.`S#` = sc.`S#`
   and student.`S#`
   in(

       select distinct s1.`S#`
       from sc s1
       where s1.score >(
           select s2.score
           from sc s2
           where s2.`S#` = 1
           and s1.`C#`= s2.`C#`
           )
   )


12) 同上
13) update SC set score=(select avg(SC_2.score) from SC SC_2 where SC_2.C#=SC.C# )
     where SC.C# in(select Course.C# from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='王志伟');

     update SC set score=(select avg(SC_2.score) from SC SC_2 where SC_2.C#=SC.C# )
     where exists(select * from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='王志伟');

*******14) select S# from SC where C# in (select C# from SC where S#='98606')
     group by S# having count(C#)=(select count(*) from SC where S#='98606');

15) delete
     from SC
     where exists(select * from course ,Teacher where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='李和');

     delete
     from SC
     where SC.C# in(select course.C# from course ,Teacher where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='李和');

16) Insert into SC
                select S#,'C602',(Select avg(score) from SC where C#='C602') from Student where S# not in (Select S# from SC where C#='C602');

     Insert into SC
                select S#,'C605',(Select avg(score) from SC where C#='C605') from Student where not exists (Select S# from SC where S#=Student.S# and C#='C605');

17) SELECT S# as 学生ID
         ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='C604') AS 编译原理
         ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='C601') AS 高等数学
         ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='C603') AS 操作系统
         --,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩
         ,COUNT(DISTINCT C#) AS 有效课程数, AVG(t.score) AS 平均成绩
     FROM SC AS t
     GROUP BY S#
     ORDER BY avg(t.score) desc

       补：

       select s_1.`S#`学号 ,
           (select sc.score from sc where sc.`C#`= 1 and sc.`S#` = s_1.`S#`) as 数据库,
           (select sc.score from sc where sc.`C#`= 2 and sc.`S#` = s_1.`S#`) as 企业管理,
           (select sc.score from sc where sc.`C#`= 3 and sc.`S#` = s_1.`S#`) as 英语,
           (select sc.score from sc where sc.`C#`= 4 and sc.`S#` = s_1.`S#`) as Java,
           (select sc.score from sc where sc.`C#`= 5 and sc.`S#` = s_1.`S#`) as C,
           count(s_1.`C#`) as 有效课程数,
           avg(s_1.score) as 平均分
           from sc s_1
           group by s_1.`S#`
           order by 平均分 desc
18) SELECT L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分
     FROM SC L ,SC AS R
     WHERE L.C# = R.C# and
         L.score = (SELECT MAX(IL.score)
                       FROM SC AS IL
                       WHERE L.C# = IL.C#
                       GROUP BY IL.C#)
         AND
         R.Score = (SELECT MIN(IR.score)
                       FROM SC AS IR
                       WHERE R.C# = IR.C#
                   GROUP BY IR.C#
                     );


                       补：

                       select s_1.`C#` 课程ID ,
                           Max(s_1.score) as 最高分,
                           Min(s_1.score) as 最低分
                           from sc s_1
                           group by s_1.`C#`
19) SELECT t.C# AS 课程号,course.Cname AS 课程名,coalesce(AVG(score),0) AS 平均成绩
         ,100 * SUM(CASE WHEN coalesce(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数
     FROM SC t,Course
     where t.C#=course.C#
     GROUP BY t.C#,course.Cname
     ORDER BY 100 * SUM(CASE WHEN coalesce(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC

   补：
      SELECT t.`C#` AS 课程号,course.Cname AS 课程名,coalesce(AVG(score),0) AS 平均成绩
         ,100 * SUM(CASE WHEN coalesce(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数
           FROM SC t,Course
           where t.`C#`=course.`C#`
           GROUP BY t.`C#`
           ORDER BY 及格百分数 DESC
20) SELECT SUM(CASE WHEN C# ='C601' THEN score ELSE 0 END)/SUM(CASE C# WHEN 'C601' THEN 1 ELSE 0 END) AS 高等数学平均分
         ,100 * SUM(CASE WHEN C# = 'C601' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = 'C601' THEN 1 ELSE 0 END) AS 高等数学及格百分数
         ,SUM(CASE WHEN C# = 'C602' THEN score ELSE 0 END)/SUM(CASE C# WHEN 'C602' THEN 1 ELSE 0 END) AS 数据结构平均分
         ,100 * SUM(CASE WHEN C# = 'C602' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = 'C602' THEN 1 ELSE 0 END) AS 数据结构及格百分数
         ,SUM(CASE WHEN C# = 'C603' THEN score ELSE 0 END)/SUM(CASE C# WHEN 'C603' THEN 1 ELSE 0 END) AS 操作系统平均分
         ,100 * SUM(CASE WHEN C# = 'C603' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = 'C603' THEN 1 ELSE 0 END) AS 操作系统及格百分数
         ,SUM(CASE WHEN C# = 'C604' THEN score ELSE 0 END)/SUM(CASE C# WHEN 'C604' THEN 1 ELSE 0 END) AS 编译原理平均分
         ,100 * SUM(CASE WHEN C# = 'C604' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = 'C604' THEN 1 ELSE 0 END) AS 编译原理及格百分数
     FROM SC

     SELECT (select avg(score) from SC where C# = 'C601') AS 高等数学平均分
         ,100 * SUM(CASE WHEN C# = 'C601' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = 'C601' THEN 1 ELSE 0 END) AS 高等数学及格百分数
         ,(select avg(score) from SC where C# = 'C602') AS 数据结构平均分
         ,100 * SUM(CASE WHEN C# = 'C602' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = 'C602' THEN 1 ELSE 0 END) AS 数据结构及格百分数
         ,(select avg(score) from SC where C# = 'C603') AS 操作系统平均分
         ,100 * SUM(CASE WHEN C# = 'C603' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = 'C603' THEN 1 ELSE 0 END) AS 操作系统及格百分数
         ,(select avg(score) from SC where C# = 'C604') AS 编译原理平均分
         ,100 * SUM(CASE WHEN C# = 'C604' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = 'C604' THEN 1 ELSE 0 END) AS 编译原理及格百分数
     FROM SC

   补：

   SELECT SUM(CASE WHEN `C#` =1 THEN score ELSE 0 END)/SUM(CASE `C#` WHEN 1 THEN 1 ELSE 0 END) AS 高等数学平均分
         ,100 * SUM(CASE WHEN `C#` =1 AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = 1 THEN 1 ELSE 0 END) AS 高等数学及格百分数
         ,SUM(CASE WHEN `C#` = 2 THEN score ELSE 0 END)/SUM(CASE `C#` WHEN 2 THEN 1 ELSE 0 END) AS 数据结构平均分
         ,100 * SUM(CASE WHEN `C#` =2 AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = 2 THEN 1 ELSE 0 END) AS 数据结构及格百分数
         ,SUM(CASE WHEN `C#` = 3 THEN score ELSE 0 END)/SUM(CASE `C#` WHEN 3 THEN 1 ELSE 0 END) AS 操作系统平均分
         ,100 * SUM(CASE WHEN `C#` = 3 AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = 3 THEN 1 ELSE 0 END) AS 操作系统及格百分数
         ,SUM(CASE WHEN `C#` = 4 THEN score ELSE 0 END)/SUM(CASE `C#` WHEN 4 THEN 1 ELSE 0 END) AS 编译原理平均分
         ,100 * SUM(CASE WHEN `C#` = 4 AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = 4 THEN 1 ELSE 0 END) AS 编译原理及格百分数
   ,sum(case when `C#`= 5 then score else 0 end)/sum(case `C#` when 5 then 1 else 0 end ) as 英语评价分
   ,100 * sum(case when `C#` = 5 and score >=60 then 1 else 0 end)/sum(case when `C#` = 5 then 1 else 0 end ) as 英语及格百分数
     FROM SC

21)、查询不同老师所教不同课程平均分从高到低显示
*****22)、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（001），自然辩证法（002），UML （003），数据库（004）
     [学生ID],[学生姓名],企业管理,自然辩证法,UML,数据库,平均成绩
23)、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
24)、查询学生平均成绩及其名次
25)、查询各科成绩前三名的记录:(不考虑成绩并列情况)
26)、查询每门课程被选修的学生数
27)、查询出只选修了一门课程的全部学生的学号和姓名
28)、查询男生、女生人数
29)、查询姓“张”的学生名单
30)、查询同名同姓学生名单，并统计同名人数

21)SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ＩＤ,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
     FROM SC AS T,Course AS C ,Teacher AS Z
     where T.C#=C.C# and C.T#=Z.T#
   GROUP BY C.C#
   ORDER BY AVG(Score) DESC
*****22)SELECT DISTINCT top 3
       SC.S# As 学生学号,
         Student.Sname AS 学生姓名 ,
       T1.score AS 企业管理,
       T2.score AS 自然辩证法,
       T3.score AS UML,
       T4.score AS 数据库,
       coalesce(T1.score,0) + coalesce(T2.score,0) + coalesce(T3.score,0) + coalesce(T4.score,0) as 总分
       FROM Student,SC LEFT JOIN SC AS T1
                       ON SC.S# = T1.S# AND T1.C# = 'C601'
             LEFT JOIN SC AS T2
                       ON SC.S# = T2.S# AND T2.C# = 'C602'
             LEFT JOIN SC AS T3
                       ON SC.S# = T3.S# AND T3.C# = 'C603'
             LEFT JOIN SC AS T4
                       ON SC.S# = T4.S# AND T4.C# = 'C604'
       WHERE student.S#=SC.S# and
       coalesce(T1.score,0) + coalesce(T2.score,0) + coalesce(T3.score,0) + coalesce(T4.score,0)
       NOT IN
       (SELECT
             DISTINCT
             TOP 15 WITH TIES
             coalesce(T1.score,0) + coalesce(T2.score,0) + coalesce(T3.score,0) + coalesce(T4.score,0)
       FROM sc
             LEFT JOIN sc AS T1
                       ON sc.S# = T1.S# AND T1.C# = 'C601'
             LEFT JOIN sc AS T2
                       ON sc.S# = T2.S# AND T2.C# = 'C602'
             LEFT JOIN sc AS T3
                       ON sc.S# = T3.S# AND T3.C# = 'C603'
             LEFT JOIN sc AS T4
                       ON sc.S# = T4.S# AND T4.C# = 'C604'
       ORDER BY coalesce(T1.score,0) + coalesce(T2.score,0) + coalesce(T3.score,0) + coalesce(T4.score,0) DESC);

   补：mysql
           SELECT DISTINCT
           SC.`S#` As 学生学号,
           Student.Sname AS 学生姓名 ,
           T1.score AS 企业管理,
           T2.score AS 自然辩证法,
           T3.score AS UML,
           T4.score AS 数据库,
           coalesce(T1.score,0) + coalesce(T2.score,0) + coalesce(T3.score,0) + coalesce(T4.score,0) as 总分
           FROM Student,
             SC LEFT JOIN SC AS T1
                       ON SC.`S#` = T1.`S#` AND T1.`C#` = 1
             LEFT JOIN SC AS T2
                       ON SC.`S#` = T2.`S#` AND T2.`C#` = 2
             LEFT JOIN SC AS T3
                       ON SC.`S#` = T3.`S#` AND T3.`C#` = 3
             LEFT JOIN SC AS T4
                       ON SC.`S#` = T4.`S#` AND T4.`C#` = 4
        WHERE student.`S#`=SC.`S#` and
        coalesce(T1.score,0) + coalesce(T2.score,0) + coalesce(T3.score,0) + coalesce(T4.score,0)
        order by 总分 desc limit 2,5




    --查询所有同学的成绩和总分，按总分降序排列，并标出序号
       select 1+(SELECT COUNT(*)
          from(
               SELECT SC.S# As 学生学号,
                 Student.Sname AS 学生姓名 ,
               T1.score AS 企业管理,
               T2.score AS 自然辩证法,
               T3.score AS UML,
               T4.score AS 数据库,
               coalesce(T1.score,0) + coalesce(T2.score,0) + coalesce(T3.score,0) + coalesce(T4.score,0) as 总分
               FROM Student,SC LEFT JOIN SC AS T1
                               ON SC.S# = T1.S# AND T1.C# = 'C601'
                     LEFT JOIN SC AS T2
                               ON SC.S# = T2.S# AND T2.C# = 'C602'
                     LEFT JOIN SC AS T3
                               ON SC.S# = T3.S# AND T3.C# = 'C603'
                     LEFT JOIN SC AS T4
                               ON SC.S# = T4.S# AND T4.C# = 'C604'
               WHERE student.S#=SC.S#
               group by SC.S#,Student.Sname,T1.score,T2.score, T3.score,T4.score
               ) as s1
             where s1.总分>s2.总分) as 名次,
               s2.学生学号,
                 s2.学生姓名 ,
               s2.企业管理,
               s2.自然辩证法,
               s2.UML,
               s2.数据库,
               s2.总分
           from (SELECT SC.S# As 学生学号,
                 Student.Sname AS 学生姓名 ,
               T1.score AS 企业管理,
               T2.score AS 自然辩证法,
               T3.score AS UML,
               T4.score AS 数据库,
               coalesce(T1.score,0) + coalesce(T2.score,0) + coalesce(T3.score,0) + coalesce(T4.score,0) as 总分
               FROM Student,SC LEFT JOIN SC AS T1
                               ON SC.S# = T1.S# AND T1.C# = 'C601'
                     LEFT JOIN SC AS T2
                               ON SC.S# = T2.S# AND T2.C# = 'C602'
                     LEFT JOIN SC AS T3
                               ON SC.S# = T3.S# AND T3.C# = 'C603'
                     LEFT JOIN SC AS T4
                               ON SC.S# = T4.S# AND T4.C# = 'C604'
               WHERE student.S#=SC.S#
               group by SC.S#,Student.Sname,T1.score,T2.score, T3.score,T4.score) as s2
               order by s2.总分 desc

23)SELECT SC.C# as 课程ID, Course.Cname as 课程名称
         ,SUM(CASE WHEN SC.score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS "[100 - 85]"
         ,SUM(CASE WHEN SC.score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS "[85 - 70]"
         ,SUM(CASE WHEN SC.score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS "[70 - 60]"
         ,SUM(CASE WHEN SC.score < 60 THEN 1 ELSE 0 END) AS "[60 -]"
     FROM SC,Course
     where SC.C#=Course.C#
     GROUP BY SC.C#,Course.Cname;

   补：
   select sc.`C#`,course.Cname,
   sum(case when sc.score <= 100 and sc.score > 85 then 1 else 0 end ) as "[100-85]",
   sum(case when sc.score between 75 and 84 then 1 else 0 end ) as "[84-75]",
   sum(case when sc.score between 65 and 74 then 1 else 0 end ) as "[74-65]",
   sum(case when sc.score < 65 then 1 else 0 end ) as "[64-0]"
   from sc,course
   where sc.`C#` = course.`C#`
   group by sc.`C#`

24)SELECT 1+(SELECT COUNT( distinct 平均成绩)
               FROM (SELECT S#,AVG(score) AS 平均成绩
                       FROM SC
                   GROUP BY S#
                   ) AS T1
             WHERE 平均成绩 > T2.平均成绩) as 名次,
       S# as 学生学号,平均成绩
     FROM (SELECT S#,AVG(score) 平均成绩
             FROM SC
         GROUP BY S#
         ) AS T2
     ORDER BY 平均成绩 desc;

25)SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
       FROM SC t1
       WHERE score IN (SELECT score
               FROM SC
               WHERE t1.C#= C#
                ORDER BY score DESC
                          fetch first 3 rows only
               )
       ORDER BY t1.C#,t1.score DESC;
26)select c#,count(S#) from sc group by C#;
27)select SC.S#,Student.Sname,count(C#) AS 选课数
   from SC ,Student
   where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1;
28)Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='男';
    Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex='女'；

       select t1.男生人数,t2.女生人数
       from (Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='男') as t1,
               (Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex='女') as t2

       select count(t1.S#) as 男生人数,count(t2.S#) as 女生人数
       from Student
           left join Student as t1 on t1.S#=Student.S# and t1.ssex='男'
           left join Student as t2 on t2.S#=Student.S# and t2.ssex='女'

29)SELECT Sname FROM Student WHERE Sname like '张%';
30)select Sname,count(*) from Student group by Sname having count(*)>1;;

31)、1981年出生的学生名单
32)、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列
33)、查询平均成绩大于85的所有学生的学号、姓名和平均成绩
34)、查询课程名称为“数据库”，且分数低于60的学生姓名和分数
35)、查询所有学生的选课情况；
36)、查询任何一门课程成绩在70分以上的姓名、课程名称和分数；
37)、查询不及格的课程，并按课程号从大到小排列
38)、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名；
39)、求选了课程的学生人数
40)、查询选修“叶平”老师所授课程的学生中，成绩最高的学生姓名及其成绩

31)select Sname, CONVERT(char (11),DATEPART(year,Sage)) as age
     from student
     where CONVERT(char(11),DATEPART(year,Sage))='1981';
32)Select C#,Avg(score) from SC group by C# order by Avg(score),C# DESC ;
33)select Sname,SC.S# ,avg(score)
     from Student,SC
     where Student.S#=SC.S# group by SC.S#,Sname having    avg(score)>85;
34)Select Sname,coalesce(score,0)
     from Student,SC,Course
     where SC.S#=Student.S# and SC.C#=Course.C# and Course.Cname='数据库'and score <60;
35)SELECT SC.S#,SC.C#,Sname,Cname
     FROM SC,Student,Course
     where SC.S#=Student.S# and SC.C#=Course.C# ;
36)SELECT distinct student.S#,student.Sname,SC.C#,SC.score
     FROM student,Sc
     WHERE SC.score>=70 AND SC.S#=student.S#;
37)select c# from sc where score <60 order by C# ;
38)select SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003';
39)select count(*) from sc;
40)select Student.Sname,score
     from Student,SC,Course C,Teacher
     where Student.S#=SC.S# and SC.C#=C.C# and C.T#=Teacher.T# and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where C#=C.C# );


41)、查询各个课程及相应的选修人数
42)、查询不同课程成绩相同的学生的学号、课程号、学生成绩
43)、查询每门功成绩最好的前两名
44)、统计每门课程的学生选修人数（超过10人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，查询结果按人数降序排列，若人数相同，按课程号升序排列
45)、检索至少选修两门课程的学生学号
46)、查询全部学生都选修的课程的课程号和课程名
47)、查询没学过“叶平”老师讲授的任一门课程的学生姓名
48)、查询两门以上不及格课程的同学的学号及其平均成绩
49)、检索“004”课程分数小于60，按分数降序排列的同学学号
50)、删除“002”同学的“001”课程的成绩

41)select count(*) from sc group by C#;
42)select distinct A.S#,A.C#,B.score from SC A ,SC B where A.Score=B.Score and A.C# <>B.C# ;
43)SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
       FROM SC t1
       WHERE score IN (SELECT score
               FROM SC
               WHERE t1.C#= C#
             ORDER BY score DESC
             fetch first 2 rows only
               )
       ORDER BY t1.C#;



44)select C# as 课程号,count(*) as 人数
     from sc
     group by C#
     order by count(*) desc,c#
45)select S#
     from sc
     group by s#
     having count(*) >= 2
46)select C#,Cname
     from Course
     where exists (select count(*) from sc where Course.C#=sc.C# group by c# having count(*)=(select count(*) from student))
47)select Sname from Student where S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# and Tname='叶平');
48)select S#,avg(coalesce(score,0)) from SC where score <60 group by S# having count(*)>2
49)select S# from SC where C#='004'and score <60 order by score desc;
50)delete from Sc where S#='001'and C#='001';
问题描述：
本题用到下面三个关系表：
CARD     借书卡。   CNO 卡号，NAME 姓名，CLASS 班级
BOOKS    图书。     BNO 书号，BNAME 书名,AUTHOR 作者，PRICE 单价，QUANTITY 库存册数
BORROW   借书记录。 CNO 借书卡号，BNO 书号，RDATE 还书日期
备注：限定每人每种书只能借一本；库存册数随借书、还书而改变。
要求实现如下15个处理：
   1)．写出建立BORROW表的SQL语句，要求定义主码完整性约束和引用完整性约束。
   2)．找出借书超过5本的读者,输出借书卡号及所借图书册数。
   3)．查询借阅了"水浒"一书的读者，输出姓名及班级。
   4)．查询过期未还图书，输出借阅者（卡号）、书号及还书日期。
   5)．查询书名包括"网络"关键词的图书，输出书号、书名、作者。
   6)．查询现有图书中价格最高的图书，输出书名及作者。
   7)．查询当前借了"计算方法"但没有借"计算方法习题集"的读者，输出其借书卡号，并按卡号降序排序输出。
   8)．将"C01"班同学所借图书的还期都延长一周。
   9)．从BOOKS表中删除当前无人借阅的图书记录。
   10)．如果经常按书名查询图书信息，请建立合适的索引。
   11)．在BORROW表上建立一个触发器，完成如下功能：如果读者借阅的书名是"数据库技术及应用"，就将该读者的借阅记录保存在BORROW_SAVE表中（注ORROW_SAVE表结构同BORROW表）。
   12)．建立一个视图，显示"力01"班学生的借书信息（只要求显示姓名和书名）。
   13)．查询当前同时借有"计算方法"和"组合数学"两本书的读者，输出其借书卡号，并按卡号升序排序输出。
   14)．假定在建BOOKS表时没有定义主码，写出为BOOKS表追加定义主码的语句。
   15)．对CARD表做如下修改：
     a. 将NAME最大列宽增加到10个字符（假定原为6个字符）。
     b. 为该表增加1列NAME（系名），可变长，最大20个字符。

1). 写出建立BORROW表的SQL语句，要求定义主码完整性约束和引用完整性约束
--实现代码：
CREATE TABLE BORROW(
     CNO int FOREIGN KEY REFERENCES CARD(CNO),
     BNO int FOREIGN KEY REFERENCES BOOKS(BNO),
     RDATE datetime,
     PRIMARY KEY(CNO,BNO))

2). 找出借书超过5本的读者,输出借书卡号及所借图书册数
--实现代码：
SELECT CNO,借图书册数=COUNT(*)
FROM BORROW
GROUP BY CNO
HAVING COUNT(*)>5

3). 查询借阅了"水浒"一书的读者，输出姓名及班级
--实现代码：
SELECT * FROM CARD c
WHERE EXISTS(
     SELECT * FROM BORROW a,BOOKS b
     WHERE a.BNO=b.BNO
         AND b.BNAME=N'水浒'
         AND a.CNO=c.CNO)

4). 查询过期未还图书，输出借阅者（卡号）、书号及还书日期
--实现代码：
SELECT * FROM BORROW
WHERE RDATE<GETDATE()

5). 查询书名包括"网络"关键词的图书，输出书号、书名、作者
--实现代码：
SELECT BNO,BNAME,AUTHOR FROM BOOKS
WHERE BNAME LIKE N'%网络%'

6). 查询现有图书中价格最高的图书，输出书名及作者
--实现代码：
SELECT BNO,BNAME,AUTHOR FROM BOOKS
WHERE PRICE=(
     SELECT MAX(PRICE) FROM BOOKS)

7). 查询当前借了"计算方法"但没有借"计算方法习题集"的读者，输出其借书卡号，并按卡号降序排序输出
--实现代码：
SELECT a.CNO
FROM BORROW a,BOOKS b
WHERE a.BNO=b.BNO AND b.BNAME=N'计算方法'
     AND NOT EXISTS(
         SELECT * FROM BORROW aa,BOOKS bb
         WHERE aa.BNO=bb.BNO
             AND bb.BNAME=N'计算方法习题集'
             AND aa.CNO=a.CNO)
ORDER BY a.CNO DESC

8). 将"C01"班同学所借图书的还期都延长一周
--实现代码：
UPDATE b SET RDATE=DATEADD(Day,7,b.RDATE)
FROM CARD a,BORROW b
WHERE a.CNO=b.CNO
     AND a.CLASS=N'C01'

9). 从BOOKS表中删除当前无人借阅的图书记录
--实现代码：
DELETE A FROM BOOKS a
WHERE NOT EXISTS(
     SELECT * FROM BORROW
     WHERE BNO=a.BNO)

10)、如果经常按书名查询图书信息，请建立合适的索引
--实现代码：
CREATE CLUSTERED INDEX IDX_BOOKS_BNAME ON BOOKS(BNAME)

11)、在BORROW表上建立一个触发器，完成如下功能：如果读者借阅的书名是"数据库技术及应用"，就将该读者的借阅记录保存在BORROW_SAVE表中（注ORROW_SAVE表结构同BORROW表）
--实现代码：
CREATE TRIGGER TR_SAVE ON BORROW
FOR INSERT,UPDATE
AS
IF @@ROWCOUNT>0
INSERT BORROW_SAVE SELECT i.*
FROM INSERTED i,BOOKS b
WHERE i.BNO=b.BNO
     AND b.BNAME=N'数据库技术及应用'

12)、建立一个视图，显示"力01"班学生的借书信息（只要求显示姓名和书名）
--实现代码：
CREATE VIEW V_VIEW
AS
SELECT a.NAME,b.BNAME
FROM BORROW ab,CARD a,BOOKS b
WHERE ab.CNO=a.CNO
     AND ab.BNO=b.BNO
     AND a.CLASS=N'力01'

13)、查询当前同时借有"计算方法"和"组合数学"两本书的读者，输出其借书卡号，并按卡号升序排序输出
--实现代码：
SELECT a.CNO
FROM BORROW a,BOOKS b
WHERE a.BNO=b.BNO
     AND b.BNAME IN(N'计算方法',N'组合数学')
GROUP BY a.CNO
HAVING COUNT(*)=2
ORDER BY a.CNO DESC

14)、假定在建BOOKS表时没有定义主码，写出为BOOKS表追加定义主码的语句
--实现代码：
ALTER TABLE BOOKS ADD PRIMARY KEY(BNO)

15.1)、将NAME最大列宽增加到10个字符（假定原为6个字符）
--实现代码：
ALTER TABLE CARD ALTER COLUMN NAME varchar(10)

15.2)、为该表增加1列NAME（系名），可变长，最大20个字符
--实现代码：
ALTER TABLE CARD ADD 系名 varchar(20)

问题描述：
为管理岗位业务培训信息，建立3个表:
S (S#,SN,SD,SA)   S#,SN,SD,SA 分别代表学号、学员姓名、所属单位、学员年龄
C (C#,CN )        C#,CN       分别代表课程编号、课程名称
SC ( S#,C#,G )    S#,C#,G     分别代表学号、所选修的课程编号、学习成绩

要求实现如下5个处理：
   1)、使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名
   2)、使用标准SQL嵌套语句查询选修课程编号为’C2’的学员姓名和所属单位
   3)、使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位
   4)、使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位
   5)、查询选修了课程的学员人数
   6)、查询选修课程超过5门的学员学号和所属单位

1)、使用标准SQL嵌套语句查询选修课程名称为’税收基础’的学员学号和姓名
--实现代码：
SELECT SN,SD FROM S
WHERE [S#] IN(
     SELECT [S#] FROM C,SC
     WHERE C.[C#]=SC.[C#]
         AND CN=N'税收基础')

2)、使用标准SQL嵌套语句查询选修课程编号为’C2’的学员姓名和所属单位
--实现代码：
SELECT S.SN,S.SD FROM S,SC
WHERE S.[S#]=SC.[S#]
     AND SC.[C#]='C2'

3)、使用标准SQL嵌套语句查询不选修课程编号为’C5’的学员姓名和所属单位
--实现代码：
SELECT SN,SD FROM S
WHERE [S#] NOT IN(
     SELECT [S#] FROM SC
     WHERE [C#]='C5')

4)、使用标准SQL嵌套语句查询选修全部课程的学员姓名和所属单位
--实现代码：
SELECT SN,SD FROM S
WHERE [S#] IN(
     SELECT [S#] FROM SC
RIGHT JOIN C ON SC.[C#]=C.[C#]
GROUP BY [S#]
HAVING COUNT(*)=COUNT(DISTINCT [S#]))
5)、查询选修了课程的学员人数
--实现代码：
SELECT 学员人数=COUNT(DISTINCT [S#]) FROM SC
6)、查询选修课程超过5门的学员学号和所属单位
--实现代码：
SELECT SN,SD FROM S
WHERE [S#] IN(
SELECT [S#] FROM SC
    GROUP BY [S#]
    HAVING COUNT(DISTINCT [C#])>5)

if not object_id('cj')is null
   drop table cj
go
create table cj(stuName nvarchar(10),KCM nvarchar(10),cj numeric(5,2))
insert into cj select '张三','语文',98
union select '李四','语文',89
union select '王五','语文',67
union select '周攻','语文',56
union select '张三','数学',89
union select '李四','数学',78
union select '王五','数学',90
union select '周攻','数学',87
方法一：
select stuname from
) cnt from cj a) x (select stuName,kcm,(select count(*) from cj where stuname!=a.stuname and kcm=a.kcm and cj>a.cj
    group by stuname having max(cnt)<=1
go
方法二：
SELECT stuname FROM cj1 a
where cj IN(SELECT TOP 2 cj FROM cj1 WHERE kcm=a.kcm ORDER BY cj desc)
GROUP BY stuname HAVING(count(1)>1)
方法三：
select distinct stuname from cj a
    where not exists(select kcm from cj b where a.stuname=stuname
                and (select count(*) from cj where kcm=b.kcm and stuname!=a.stuname and cj>b.cj)>1)