sql练习

2016.3.14

原帖链接:http://bbs.csdn.net/topics/280002741

表架构

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

建表语句

CREATE TABLE student 
  ( 
     s#    INT, 
     sname nvarchar(32), 
     sage  INT, 
     ssex  nvarchar(8) 
  ) 

CREATE TABLE course 
  ( 
     c#    INT, 
     cname nvarchar(32), 
     t#    INT 
  ) 

CREATE TABLE sc 
  ( 
     s#    INT, 
     c#    INT, 
     score INT 
  ) 

CREATE TABLE teacher 
  ( 
     t#    INT, 
     tname nvarchar(16) 
  )

插入测试数据语句

insert into Student select 1,N'刘一',18,N'男' union all
 select 2,N'钱二',19,N'女' union all
 select 3,N'张三',17,N'男' union all
 select 4,N'李四',18,N'女' union all
 select 5,N'王五',17,N'男' union all
 select 6,N'赵六',19,N'女' 
 
 insert into Teacher select 1,N'叶平' union all
 select 2,N'贺高' union all
 select 3,N'杨艳' union all
 select 4,N'周磊'
 
 insert into Course select 1,N'语文',1 union all
 select 2,N'数学',2 union all
 select 3,N'英语',3 union all
 select 4,N'物理',4
 
 insert into SC 
 select 1,1,56 union all 
 select 1,2,78 union all 
 select 1,3,67 union all 
 select 1,4,58 union all 
 select 2,1,79 union all 
 select 2,2,81 union all 
 select 2,3,92 union all 
 select 2,4,68 union all 
 select 3,1,91 union all 
 select 3,2,47 union all 
 select 3,3,88 union all 
 select 3,4,56 union all 
 select 4,2,88 union all 
 select 4,3,90 union all 
 select 4,4,93 union all 
 select 5,1,46 union all 
 select 5,3,78 union all 
 select 5,4,53 union all 
 select 6,1,35 union all 
 select 6,2,68 union all 
 select 6,4,71

问题

1、查询“001”课程比“002”课程成绩高的所有学生的学号； 
  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#; 
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； 
  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 
4、查询姓“李”的老师的个数； 
  select count(distinct(Tname)) 
  from Teacher 
  where Tname like '李%'; 
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； 
  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； 
  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')); 
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； 
  Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2 
  from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2 <score; 
9、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# not in (select S.S# from Student AS S,SC where S.S#=SC.S# and score>60); 
10、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#) <(select count(C#) from Course); 
11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名； 
    select distinct S#,Sname from Student,SC where Student.S#=SC.S# and SC.C# in (select C# from SC where S#='1001'); 
12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名； 
    select distinct SC.S#,Sname 
    from Student,SC 
    where Student.S#=SC.S# and C# in (select C# from SC where S#='001'); 
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； 
    update SC set score=(select avg(SC_2.score) 
    from SC SC_2 
    where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='叶平'); 
14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名； 
   SELECT
　　student.s#,
　　student.sname
　　FROM
　　sc,
student
WHERE
sc.s# = student.s#
AND student.s NOT IN (
SELECT
sc.s#
FROM
sc
WHERE
sc.c# NOT IN (SELECT sc.c# FROM sc WHERE sc.s# = 2)
)
AND sc.s# != 2
GROUP BY
student.s#,
student.sname
HAVING
COUNT(*) = (
SELECT
COUNT(*)
FROM
sc
WHERE
sc.s# = 2
);
15、删除学习“叶平”老师课的SC表记录； 
    Delect SC 
    from course ,Teacher  
    where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平'; 
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“003”课程的同学学号、2、 
    号课的平均成绩； 
    Insert SC select S#,'002',(Select avg(score) 
    from SC where C#='002') from Student where S# not in (Select S# from SC where C#='002'); 
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 
    SELECT S# as 学生ID 
        ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库 
        ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理 
        ,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语 
        ,COUNT(*) 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)  
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 
    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,Student AS IM 
                      WHERE L.C# = IL.C# and IM.S#=IL.S# 
                      GROUP BY IL.C#) 
        AND 
        R.Score = (SELECT MIN(IR.score) 
                      FROM SC AS IR 
                      WHERE R.C# = IR.C# 
                  GROUP BY IR.C# 
                    ); 
自己写的:select c# ,max(score)as 最高分 ,min(score) as 最低分 from dbo.sc  group by c#
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 
    SELECT t.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩 
        ,100 * SUM(CASE WHEN  isnull(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 100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC 
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），马克思（002），OO&UML （003），数据库（004） 
    SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分 
        ,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数 
        ,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分 
        ,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数 
        ,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分 
        ,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数 
        ,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分 
        ,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数 
  FROM SC
21、查询不同老师所教不同课程平均分从高到低显示 
  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、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），马克思（002），UML （003），数据库（004） 
    [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩 
    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 马克思, 
      T3.score AS UML, 
      T4.score AS 数据库, 
      ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分 
      FROM Student,SC  LEFT JOIN SC AS T1 
                      ON SC.S# = T1.S# AND T1.C# = '001' 
            LEFT JOIN SC AS T2 
                      ON SC.S# = T2.S# AND T2.C# = '002' 
            LEFT JOIN SC AS T3 
                      ON SC.S# = T3.S# AND T3.C# = '003' 
            LEFT JOIN SC AS T4 
                      ON SC.S# = T4.S# AND T4.C# = '004' 
      WHERE student.S#=SC.S# and 
      ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) 
      NOT IN 
      (SELECT 
            DISTINCT 
            TOP 15 WITH TIES 
            ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) 
      FROM sc 
            LEFT JOIN sc AS T1 
                      ON sc.S# = T1.S# AND T1.C# = 'k1' 
            LEFT JOIN sc AS T2 
                      ON sc.S# = T2.S# AND T2.C# = 'k2' 
            LEFT JOIN sc AS T3 
                      ON sc.S# = T3.S# AND T3.C# = 'k3' 
            LEFT JOIN sc AS T4 
                      ON sc.S# = T4.S# AND T4.C# = 'k4' 
      ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC); 

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

24、查询学生平均成绩及其名次 
      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、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#; 
26、查询每门课程被选修的学生数 
  select c#,count(S#) from sc group by C#; 
27、查询出只选修了一门课程的全部学生的学号和姓名 
  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、查询男生、女生人数 
    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'； 
29、查询姓“张”的学生名单 
    SELECT Sname FROM Student WHERE Sname like '张%'; 
30、查询同名同性学生名单，并统计同名人数 
  select Sname,count(*) from Student group by Sname having  count(*)>1;; 
31、1981年出生的学生名单(注：Student表中Sage列的类型是datetime) 
    select Sname,  CONVERT(char (11),DATEPART(year,Sage)) as age 
    from student 
    where  CONVERT(char(11),DATEPART(year,Sage))='1981'; 
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 
    Select C#,Avg(score) from SC group by C# order by Avg(score),C# DESC ; 
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 
    select Sname,SC.S# ,avg(score) 
    from Student,SC 
    where Student.S#=SC.S# group by SC.S#,Sname having    avg(score)>85; 
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 
    Select Sname,isnull(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、查询所有学生的选课情况； 
    SELECT SC.S#,SC.C#,Sname,Cname 
    FROM SC,Student,Course 
    where SC.S#=Student.S# and SC.C#=Course.C# ; 
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； 
    SELECT  distinct student.S#,student.Sname,SC.C#,SC.score 
    FROM student,Sc 
    WHERE SC.score>=70 AND SC.S#=student.S#; 
37、查询不及格的课程，并按课程号从大到小排列 
    select c# from sc where scor e <60 order by C# ; 
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； 
    select SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003'; 
39、求选了课程的学生人数 
    select count(*) from sc; 
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 
    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 
    select count(*) from sc group by C#; 
42、查询不同课程成绩相同的学生的学号、课程号、学生成绩 
  select distinct  A.S#,B.score from SC A  ,SC B where A.Score=B.Score and A.C# <>B.C# ; 
43、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#; 
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  
    select  C# as 课程号,count(*) as 人数 
    from  sc  
    group  by  C# 
    order  by  count(*) desc,c#  
45、检索至少选修两门课程的学生学号 
    select  S#  
    from  sc  
    group  by  s# 
    having  count(*)  >  =  2 
46、查询全部学生都选修的课程的课程号和课程名 
    select  C#,Cname  
    from  Course  
    where  C#  in  (select  c#  from  sc group  by  c#)  
47、查询没学过“叶平”老师讲授的任一门课程的学生姓名 
    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、查询两门以上不及格课程的同学的学号及其平均成绩 
    select S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where score <60 group by S# having count(*)>2)group by S#; 
49、检索“004”课程分数小于60，按分数降序排列的同学学号 
    select S# from SC where C#='004'and score <60 order by score desc; 
50、删除“002”同学的“001”课程的成绩 
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
--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 
--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)

----------------------------------------------------------------------分界线------------------------------------------------------------------------------------

--
-- select * from t_person1
--
--insert into t_person1 values ('jason',1,18)


--create table t_students
--(
--studentid int,
--iname nvarchar(20),
--courseID nvarchar(20),
--score int
--)
--
--create table  t_course
--(
--courseID int,
--courseName nvarchar(20),
--teacherID int
--)
--
--create table t_teacher
--(
--teacherID int,
--teacherName nvarchar(20)
--)

--
--select * from dbo.t_course
--select * from  dbo.t_students
--select * from  dbo.t_teacher
--
--insert into dbo.t_course values (1,'物理',5)
--insert into dbo.t_course values (2,'化学',3)
--insert into dbo.t_course values (3,'数学',2)
--insert into dbo.t_course values (4,'语文',4)
--insert into dbo.t_course values (6,'英语',1)
--insert into dbo.t_course values (5,'历史',6)
--
--insert into dbo.t_teacher values (1,'张老师')
--insert into dbo.t_teacher values (2,'赵老师')
--insert into dbo.t_teacher values (3,'钱老师')
--insert into dbo.t_teacher values (4,'孙老师')
--insert into dbo.t_teacher values (5,'李老师')
--insert into dbo.t_teacher values (6,'周老师')
--
-- 
--
--insert into dbo.t_students values (1,'赵同学',2,88)
--insert into dbo.t_students values (1,'赵同学',4,20)
--insert into dbo.t_students values (1,'赵同学',5,54)
--insert into dbo.t_students values (2,'钱同学',1,86)
--insert into dbo.t_students values (2,'钱同学',2,55)
--insert into dbo.t_students values (2,'钱同学',5,44)
--insert into dbo.t_students values (3,'孙同学',2,58)
--insert into dbo.t_students values (3,'孙同学',3,33)
--insert into dbo.t_students values (3,'孙同学',4,58)
--insert into dbo.t_students values (4,'李同学',2,69)
--insert into dbo.t_students values (5,'周同学',6,66)
--insert into dbo.t_students values (6,'吴同学',6,100)

 
--选修了化学的同学的名字，和分数
--化学  李同学  48
--solution 1
--select '化学',iname,score
--from t_students
--where courseID =
--(
--select courseID from dbo.t_course where courseName = '化学'
--)

--solution 2
--select c.courseName,s.iname,s.score
--from t_students as s
--left join t_course as c
--on s.courseID = c.courseID
--where c.courseName = '化学'

--选修了英语的并且分数大于80分的同学名字，任课老师名字
--英语 赵同学 30 李老师
--t_students as s
--t_course as c
--t_teacher as t
--courseName  iname   score   teacherName
select c.courseName,s.iname,s.score,t.teacherName
from t_students as s
left join t_course as c on s.courseID = c.courseID
left join t_teacher as t on c.teacherID = t.teacherID
where c.courseName = '英语'
and s.score > 80

2016-1-9

今天在网上找了几道经典的SQL练习题做了一下，虽然都不难，但是对打基础是很有好处的，在明白的基础上可以进一步做分析，来研究一下各种解法的优劣，甚至进行简单的优化。。
现在将题目和答案分享一下。我使用的是MYSQL 5.0，但是绝大部分都是标准SQL。
表结构：

CREATE TABLE STUDENT
(SNO VARCHAR(3) NOT NULL,
SNAME VARCHAR(4) NOT NULL,
SSEX VARCHAR(2) NOT NULL,
SBIRTHDAY DATETIME,
CLASS VARCHAR(5))
go
CREATE TABLE COURSE
(CNO VARCHAR(5) NOT NULL,
CNAME VARCHAR(10) NOT NULL,
TNO VARCHAR(10) NOT NULL)
go
CREATE TABLE SCORE
(SNO VARCHAR(3) NOT NULL,
CNO VARCHAR(5) NOT NULL,
DEGREE NUMERIC(10, 1) NOT NULL)
go
CREATE TABLE TEACHER
(TNO VARCHAR(3) NOT NULL,
TNAME VARCHAR(4) NOT NULL, TSEX VARCHAR(2) NOT NULL,
TBIRTHDAY DATETIME NOT NULL, PROF VARCHAR(6),
DEPART VARCHAR(10) NOT NULL)

INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (108 ,'曾华'
,'男' ,1977-09-01,95033);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (105 ,'匡明'
,'男' ,1975-10-02,95031);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (107 ,'王丽'
,'女' ,1976-01-23,95033);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (101 ,'李军'
,'男' ,1976-02-20,95033);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (109 ,'王芳'
,'女' ,1975-02-10,95031);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (103 ,'陆君'
,'男' ,1974-06-03,95031);
GO
INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('3-105' ,'计算机导论',825)
INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('3-245' ,'操作系统' ,804);
INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('6-166' ,'数据电路' ,856);
INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('9-888' ,'高等数学' ,100);
GO
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (103,'3-245',86);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (105,'3-245',75);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (109,'3-245',68);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (103,'3-105',92);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (105,'3-105',88);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (109,'3-105',76);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (101,'3-105',64);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (107,'3-105',91);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (108,'3-105',78);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (101,'6-166',85);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (107,'6-106',79);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (108,'6-166',81);
GO
INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
VALUES (804,'李诚','男','1958-12-02','副教授','计算机系');
INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
VALUES (856,'张旭','男','1969-03-12','讲师','电子工程系');
INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
VALUES (825,'王萍','女','1972-05-05','助教','计算机系');
INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
VALUES (831,'刘冰','女','1977-08-14','助教','电子工程系');

2016-1-9（未完待续，做到17题）
2016-1-11（未完待续，做到25题,18,21题不会做）
2016-4-25全部做完

--1、 查询Student表中的所有记录的Sname、Ssex和Class列。
select Sname,Ssex,Class  from dbo.STUDENT
--2、 查询教师所有的单位即不重复的Depart列。
select distinct Depart from dbo.TEACHER
--3、 查询Student表的所有记录。
select * from dbo.STUDENT
--4、 查询Score表中成绩在60到80之间的所有记录。
select * from Score where degree between 60 and 80
--5、 查询Score表中成绩为85，86或88的记录。
select * from Score where degree in (85,86,88)
--6、 查询Student表中“95031”班或性别为“女”的同学记录。
select * from dbo.STUDENT where class = '95031' or ssex = '女'
--7、 以Class降序查询Student表的所有记录。
select * from dbo.STUDENT order by class desc
--8、 以Cno升序、Degree降序查询Score表的所有记录。
select * from dbo.SCORE order by cno asc,degree desc
--9、 查询“95031”班的学生人数。
select count(1) from dbo.STUDENT where class = '95031'
--10、查询Score表中的最高分的学生学号和课程号。
select sno,cno from score where degree = (select max(degree) from score)
--11、查询‘3-105’号课程的平均分。
select avg(degree) from score where cno = '3-105'
--12、查询Score表中至少有5名学生选修的并以3开头的课程的平均分数。
select avg(degree) from score group by cno having (count(1)>5 and cno like '3%')
--13、查询最低分大于70，最高分小于90的Sno列。
select sno from dbo.SCORE group by sno having min(degree)>70 and max(degree)<90
--14、查询所有学生的Sname、Cno和Degree列。
select (select sname from student where sno = sc.sno) sname ,cno ,degree from score sc
--15、查询所有学生的Sno、Cname和Degree列。
select sno,(select cname from course where cno = sc.cno)cname,degree from score sc
--16、查询所有学生的Sname、Cname和Degree列。
select
(select sname from student where sno = sc.sno)sname,
(select cname from course where cno = sc.cno)cname,
degree from score sc
--17、查询“95033”班所选课程的平均分。
select avg(degree) from score where sno in( select sno from student where class = '95033')
--这一句为什么错--select avg(degree) from score where exists(select sno from student where class = '95033')
--18、假设使用如下命令建立了一个grade表：
--create table GRADE
--(
--low   numeric(3,0),
--upp   numeric(3),
--rank   char(1)
--)
--insert into grade values(90,100,'A');
--insert into grade values(80,89,'B');
--insert into grade values(70,79,'C');
--insert into grade values(60,69,'D');
--insert into grade values(0,59,'E');
--现查询所有同学的Sno、Cno和rank列。
select s.sno,s.cno, 
(select rank from grade as g where s.degree between g.low and g.upp)as rank
from dbo.SCORE as s
--19、查询选修“3-105”课程的成绩高于“109”号同学成绩的所有同学的记录。
select * from score where cno = '3-105'
and degree > (select degree from score where sno = '109' and cno = '3-105')
--20、查询score中选学一门以上课程的同学中分数为非最高分成绩的记录。
SELECT * FROM score s WHERE
DEGREE<(SELECT MAX(DEGREE) FROM SCORE) GROUP BY SNO HAVING COUNT(SNO)>1 ORDER BY DEGREE
--21、查询成绩高于学号为“109”、课程号为“3-105”的成绩的所有记录。
select * from dbo.SCORE where DEGREE >
(select DEGREE from dbo.SCORE where SNO='109' and CNO = '3-105')
--22、查询和学号为108的同学同年出生的所有学生的Sno、Sname和Sbirthday列。
select sno,sname,sbirthday from student where
datediff(year,sbirthday,(select sbirthday from student where sno = '108')) = 0
--23、查询“张旭“教师任课的学生成绩。
select * from score where cno = (select cno from course where tno = (select tno from teacher where tname = '张旭'))
--24、查询选修某课程的同学人数多于5人的教师姓名。
select tname from teacher where
tno = (select tno from course where cno =  
( select cno from score group by cno having count(sno) > 5 ))
--25、查询95033班和95031班全体学生的记录。
select * from student left join score on student.sno = score.sno where student.class in ('95033','95031')
--26、查询存在有85分以上成绩的课程Cno.
select distinct cno from dbo.SCORE where DEGREE >= 85
--27、查询出“计算机系“教师所教课程的成绩表。
select * from dbo.SCORE where cno in
(select cno from dbo.COURSE where tno in (select tno from dbo.TEACHER where depart = '计算机系'))
--28、查询“计算机系”与“电子工程系“不同职称的教师的Tname和Prof。
select Tname,Prof from dbo.TEACHER where prof not in
(
    select prof from dbo.TEACHER  where depart = '电子工程系' and prof in 
    (
        select prof from dbo.TEACHER  where depart = '计算机系'
    )
)
--29、查询选修编号为“3-105“课程且成绩至少高于选修编号为“3-245”的同学的Cno、Sno和Degree,并按Degree从高到低次序排序。
select * from dbo.SCORE where cno = '3-105' and DEGREE >=
any(select DEGREE from dbo.SCORE where cno = '3-245') order by DEGREE desc
--30、查询选修编号为“3-105”且成绩高于选修编号为“3-245”课程的同学的Cno、Sno和Degree.
select * from dbo.SCORE where cno = '3-105' and DEGREE >=
all(select DEGREE from dbo.SCORE where cno = '3-245')
--31、查询所有教师和同学的name、sex和birthday.
select Sname as name,ssex as sex,sbirthday as birthday,'S' as o from dbo.STUDENT
union 
select tname as name,tsex as sex,tbirthday as birthday,'T' as o from dbo.TEACHER
order by birthday asc
--32、查询所有“女”教师和“女”同学的name、sex和birthday.
select Sname as name,ssex as sex,sbirthday as birthday,'S' as o from dbo.STUDENT where ssex = '女'
union 
select tname as name,tsex as sex,tbirthday as birthday,'T' as o from dbo.TEACHER where tsex = '女'
order by birthday asc
--33、查询成绩比该课程平均成绩低的同学的成绩表。
select * from dbo.SCORE as a where DEGREE < (select avg(DEGREE) from SCORE where cno = a.cno)
--34、查询所有任课教师的Tname和Depart.
select Tname,Depart from dbo.TEACHER where tno in
(select tno from dbo.COURSE)
--35  查询所有未讲课的教师的Tname和Depart.
select Tname,Depart from dbo.TEACHER where tno not in
(select tno from dbo.COURSE)
--36、查询至少有2名男生的班号。
select class from STUDENT group by class having count(ssex) >= 2 
--37、查询Student表中不姓“王”的同学记录。
select * from dbo.STUDENT where SNAME not like '王%'
--38、查询Student表中每个学生的姓名和年龄。
select SNAME,datediff(year,SBIRTHDAY,getdate())as AGE from dbo.STUDENT
--39、查询Student表中最大和最小的Sbirthday日期值。
select max(SBIRTHDAY) as up,min(SBIRTHDAY) as low from dbo.STUDENT
--40、以班号和年龄从大到小的顺序查询Student表中的全部记录。
select * from dbo.STUDENT order by class desc,sbirthday desc
--41、查询“男”教师及其所上的课程。
select t.tname,c.cname from TEACHER as t 
right join COURSE   c
on t.tno = c.tno
where t.tsex = '男'
--42、查询最高分同学的Sno、Cno和Degree列。
select * from SCORE where DEGREE =
(select max(DEGREE) from dbo.SCORE)
--43、查询和“李军”同性别的所有同学的Sname.
select SNAME from STUDENT where ssex =
(select ssex from dbo.STUDENT where SNAME  = '李军')
--44、查询和“李军”同性别并同班的同学Sname.
select SNAME from STUDENT where ssex =
(select ssex from dbo.STUDENT where SNAME  = '李军')
and class = 
(select class from dbo.STUDENT where SNAME  = '李军')
--45、查询所有选修“计算机导论”课程的“男”同学的成绩表
select * from dbo.SCORE where cno = 
(select cno  from dbo.COURSE where cname = '计算机导论')
and sno in 
(select sno from dbo.STUDENT where ssex = '男')

参考答案：
1. SELECT SNAME,SSEX,CLASS FROM STUDENT;

2. SELECT DISTINCT DEPART FROM TEACHER;

3. SELECT * FROM STUDENT;

4. SELECT * FROM SCORE WHERE DEGREE BETWEEN 60 AND 80;

5.SELECT * FROM SCORE WHERE DEGREE IN (85,86,88);

6. SELECT * FROM STUDENT WHERE CLASS='95031' OR SSEX='女';

7.SELECT * FROM STUDENT ORDER BY CLASS DESC;

8.SELECT * FROM SCORE ORDER BY CNO ASC,DEGREE DESC;

9.SELECT COUNT(*) FROM STUDENT WHERE CLASS='95031';

10.SELECT SNO,CNO FROM SCORE WHERE DEGREE=(SELECT MAX(DEGREE) FROM SCORE);

SELECT SNO,CNO FROM SCORE ORDER BY DEGREE DESC LIMIT 1;

11.SELECT AVG(DEGREE) FROM SCORE WHERE CNO='3-105';

12.select avg(degree),cno
from score
where cno like '3%'
group by cno
having count(sno)>= 5;

13.SELECT SNO FROM SCORE GROUP BY SNO HAVING MIN(DEGREE)>70 AND MAX(DEGREE)<90;

14.SELECT A.SNAME,B.CNO,B.DEGREE FROM STUDENT AS A JOIN SCORE AS B ON A.SNO=B.SNO;

15.SELECT A.CNAME, B.SNO,B.DEGREE FROM COURSE AS A JOIN SCORE AS B ON A.CNO=B.CNO ;

16.SELECT A.SNAME,B.CNAME,C.DEGREE FROM STUDENT A JOIN (COURSE B,SCORE C)
ON A.SNO=C.SNO AND B.CNO =C.CNO;

17.SELECT AVG(A.DEGREE) FROM SCORE A JOIN STUDENT B ON A.SNO = B.SNO WHERE B.CLASS='95033';

18.SELECT A.SNO,A.CNO,B.RANK FROM SCORE A,GRADE B WHERE A.DEGREE BETWEEN B.LOW AND B.UPP

ORDER BY RANK;

19.SELECT A.* FROM SCORE A JOIN SCORE B WHERE A.CNO='3-105' AND A.DEGREE>B.DEGREE AND

B.SNO='109' AND B.CNO='3-105';
另一解法：SELECT A.* FROM SCORE A WHERE A.CNO='3-105' AND A.DEGREE>ALL(SELECT DEGREE FROM

SCORE B WHERE B.SNO='109' AND B.CNO='3-105');

20.SELECT * FROM score s WHERE DEGREE<(SELECT MAX(DEGREE) FROM SCORE) GROUP BY SNO HAVING

COUNT(SNO)>1 ORDER BY DEGREE ;

21.见19的第二种解法

22。SELECT SNO,SNAME,SBIRTHDAY FROM STUDENT WHERE YEAR(SBIRTHDAY)=(SELECT YEAR(SBIRTHDAY)

FROM STUDENT WHERE SNO='108');
ORACLE:select x.cno,x.Sno,x.degree from score x,score y where x.degree>y.degree and

y.sno='109'and y.cno='3-105';
select cno,sno,degree from score where degree >(select degree from score where sno='109'

and cno='3-105')

23.SELECT A.SNO,A.DEGREE FROM SCORE A JOIN (TEACHER B,COURSE C)
ON A.CNO=C.CNO AND B.TNO=C.TNO
WHERE B.TNAME='张旭';
另一种解法：select cno,sno,degree from score where cno=(select x.cno from course x,teacher y

where x.tno=y.tno and y.tname='张旭');
根据实际EXPLAIN此SELECT语句，第一个的扫描次数要小于第二个

24.SELECT A.TNAME FROM TEACHER A JOIN (COURSE B, SCORE C) ON (A.TNO=B.TNO AND B.CNO=C.CNO)

GROUP BY C.CNO HAVING COUNT(C.CNO)>5;
另一种解法：select tname from teacher where tno in(select x.tno from course x,score y where

x.cno=y.cno group by x.tno having count(x.tno)>5);
实际测试1明显优于2

25。select cno,sno,degree from score where cno=(select x.cno from course x,teacher y where

x.tno=y.tno and y.tname='张旭');

26。SELECT CNO FROM SCORE GROUP BY CNO HAVING MAX(DEGREE)>85;
另一种解法：select distinct cno from score where degree in (select degree from score where

degree>85);

27。SELECT A.* FROM SCORE A JOIN (TEACHER B,COURSE C) ON A.CNO=C.CNO AND B.TNO=C.TNO
WHERE B.DEPART='计算机系';
另一种解法：SELECT * from score where cno in (select a.cno from course a join teacher b on

a.tno=b.tno and b.depart='计算机系');
此时2略好于1，在多连接的境况下性能会迅速下降

28。select tname,prof from teacher where depart='计算机系' and prof not in (select prof from

teacher where depart='电子工程系');

29。SELECT * FROM SCORE WHERE DEGREE>ANY(SELECT DEGREE FROM SCORE WHERE CNO='3-245') ORDER

BY DEGREE DESC;

30。SELECT * FROM SCORE WHERE DEGREE>ALL(SELECT DEGREE FROM SCORE WHERE CNO='3-245') ORDER

BY DEGREE DESC;

31.SELECT SNAME AS NAME, SSEX AS SEX, SBIRTHDAY AS BIRTHDAY FROM STUDENT
UNION
SELECT TNAME AS NAME, TSEX AS SEX, TBIRTHDAY AS BIRTHDAY FROM TEACHER;

32.SELECT SNAME AS NAME, SSEX AS SEX, SBIRTHDAY AS BIRTHDAY FROM STUDENT WHERE SSEX='女'
UNION
SELECT TNAME AS NAME, TSEX AS SEX, TBIRTHDAY AS BIRTHDAY FROM TEACHER WHERE TSEX='女';

33.SELECT A.* FROM SCORE A WHERE DEGREE<(SELECT AVG(DEGREE) FROM SCORE B WHERE A.CNO=B.CNO);
须注意********此题

34。解法一：SELECT A.TNAME,A.DEPART FROM TEACHER A JOIN COURSE B ON A.TNO=B.TNO;
解法二：select tname,depart from teacher a where exists
(select * from course b where a.tno=b.tno);
解法三：SELECT TNAME,DEPART FROM TEACHER WHERE TNO IN (SELECT TNO FROM COURSE);

实际分析，第一种揭发貌似更好，至少扫描次数最少。

35.解法一：SELECT TNAME,DEPART FROM TEACHER A LEFT JOIN COURSE B USING(TNO) WHERE ISNUL

(B.tno);
解法二：select tname,depart from teacher a where not exists
(select * from course b where a.tno=b.tno);
解法三：SELECT TNAME,DEPART FROM TEACHER WHERE TNO NOT IN (SELECT TNO FROM COURSE);
NOT IN的方法效率最差，其余两种差不多

36.SELECT CLASS FROM STUDENT A WHERE SSEX='男' GROUP BY CLASS HAVING COUNT(SSEX)>1;

37.SELECT * FROM STUDENT A WHERE SNAME not like '王%';

38.SELECT SNAME,(YEAR(NOW())-YEAR(SBIRTHDAY)) AS AGE FROM STUDENT;

39.select sname,sbirthday as THEMAX from student where sbirthday =(select min(SBIRTHDAY)

from student)
union
select sname,sbirthday as THEMIN from student where sbirthday =(select max(SBIRTHDAY) from

student);

40.SELECT CLASS,(YEAR(NOW())-YEAR(SBIRTHDAY)) AS AGE FROM STUDENT ORDER BY CLASS DESC,AGE

DESC;

41.SELECT A.TNAME,B.CNAME FROM TEACHER A JOIN COURSE B USING(TNO) WHERE A.TSEX='男';

42.SELECT A.* FROM SCORE A WHERE DEGREE=(SELECT MAX(DEGREE) FROM SCORE B );

43.SELECT SNAME FROM STUDENT A WHERE SSEX=(SELECT SSEX FROM STUDENT B WHERE B.SNAME='李军');

44.SELECT SNAME FROM STUDENT A WHERE SSEX=(SELECT SSEX FROM STUDENT B WHERE B.SNAME='李军' )
AND CLASS=(SELECT CLASS FROM STUDENT C WHERE c.SNAME='李军');

45.解法一：SELECT A.* FROM SCORE A JOIN (STUDENT B,COURSE C) USING(sno,CNO) WHERE B.SSEX='男

' AND C.CNAME='计算机导论';
解法二：select * from score where sno in(select sno from student where
ssex='男') and cno=(select cno from course
where cname='计算机导论');

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posted @ 2015-11-27 10:05 Y档案Y 阅读(480) 评论(0) 收藏举报

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