)1，写出一条Sql语句：取出表A中第31到第40记录（SQLServer,以自动增长的ID作为主键,注意：ID可能不是连续的。
答：
→ 解1: select top 10 * from A where id not in (select top 30 id from A)
演变步骤：
1）select top 30 id from T_FilterWords–取前条
2）select * from T_FilterWords where id not in (select top 30 id from T_FilterWords)–取id不等于前三十条的–也就是把前条排除在外
3）select top 10 * from T_FilterWords where id not in (select top 30 id from T_FilterWords)
–取把前条排除在外的前条，也就是-40条
→ 解2: select top 10 * from A where id > (select max(id) from (select top 30 id from A )as A)
→ 解3：用ROW_NUMBER实现
/———————————————–+———————————————–\
下面的SQL题是常考题中的常考题，必须重视！！！！！！！！
2，横表、纵表转换（常考!!!）
1）纵表结构 TableA
NameCourseGrade张三语文75张三数学80张三英语90李四语文95李四数学55
横表结构 TableB
Name语文数学英语张三758090李四95550
先理解：
select Name，
(case Course when ‘语文‘ then Grade else 0 end) as 语文，
(case Course when ‘数学‘ then Grade else 0 end) as 数学，
(case Course when ‘英语‘ then Grade else 0 end) as 英语
from TableA
然后理解标准答案：
select Name，
sum(case Course when ‘语文‘ then Grade else 0 end) as 语文，
sum(case Course when ‘数学‘ then Grade else 0 end) as 数学，
sum(case Course when ‘英语‘ then Grade else 0 end) as 英语
from TableA
group by Name
2）、横表转纵表的”SQL”示例
横表结构: TEST_H2Z
ID 姓名 语文 数学 英语
1 张三 80 90 70
2 李四 90 85 95
3 王五 88 75 90
转换后的表结构:
ID 姓名 科目 成绩
1 张三 语文 80
2 张三 数学 90
3 张三 英语 70
4 李四 语文 90
5 李四 数学 80
6 李四 英语 99
7 王五 语文 85
8 王五 数学 96
9 王五 英语 88
横表转纵表SQL示例:
SELECT 姓名,’语文’ AS 科目,语文 AS 成绩 FROM TEST_H2Z UNION ALL
SELECT 姓名,’数学’ AS 科目,数学 AS 成绩 FROM TEST_H2Z UNION ALL
SELECT 姓名,’英语’ AS 科目,英语 AS 成绩 FROM TEST_H2Z
ORDER BY 姓名,科目 DESC;
3，删除姓名、年龄重复的记录(只保留Id最大的一条)（常考！！！）
Id name age salary
1 yzk 80 1000
2 yzk 80 2000
3 tom 20 20000
4 tom 20 20000
5 im 20 20000
//取得不重复的数据
select * from Persons
where Id in
(
SELECT MAX(Id) AS Expr1
FROM Persons
GROUP BY Name, Age
)
→ 根据姓名、年龄分组，取出每组的Id最大值，然后将Id最大值之外的排除。
删除重复的数据：
delete from Persons
where Id not in
(
SELECT MAX(Id) AS Expr1
FROM Persons
GROUP BY Name, Age
)
4，下面是一个由*号组成的4行倒三角形图案。
→ 要求：1、输入倒三角形的行数，行数的取值3-21之间，对于非法的行数，要求抛出提示“非法行数！”；2、在屏幕上打印这个指定了行数的倒三角形。
*******
*****
***
*
5，一个文本文件含有如下内容：
4580616022644994|3000|赵涛
4580616022645017|6000|张屹
4580616022645090|3200|郑欣夏
→ 上述文件每行为一个转账记录，第一列表示帐号，第二列表示金额，第三列表示开户人姓名。
创建一张数据库表（MS SQLServer数据库，表名和字段名自拟），请将上述文件逐条插入此表中。
6、一个文本文件含有如下内容，分别表示姓名和成绩：
张三 90
李四 96
王五 78
赵六 82
→ 提供用户一个控制台界面，允许用户输入要查询的姓名，输入姓名并且按回车以后，打印出此人的成绩，如果不输入姓名直接按回车则显示所有人的姓名以及成绩。（注意：不能使用数据库）
7，表A字段Id为numberic(18,0)，哪个SQL语句是错误的:
select * from A where id=”;
select * from A where id=’13′;
select * from A where id=null;
select * from A where id=’ 13′;
8，在SQLServer中求当前时间与2012-01-01 0：0:0相差的秒数？
9，做一个表格，三行三列，第一列，前两行合一；第二行，后两列合一。
10，
表一：student_info
学号姓名性别出生年月家庭住址备注0001张三男1981-8-9北京NULL
表二：curriculum
课程编号课程名称学分0001计算机基础20002C语言2
表三：grade
学号课程编号分数00010001800001000290
题目：
→ 条件查询：
在GRADE表中查找80-90份的学生学号和分数
select 学号,分数 from grade where 分数 between 80 and 90
在GRADE 表中查找课程编号为003学生的平均分
select avg(分数) from grade where 课程编号=’003′
在GRADE 表中查询学习各门课程的人数
Select课程编号,count(学号) as 人数from grade group by 课程编号
查询所有姓张的学生的学号和姓名
select 姓名,学号 from student_info where 姓名 like ‘张%’
→ 嵌套查询：
1，查询和学号’0001’的这位同学性别相同的所有同学的姓名和出生年月
select 姓名,出生年月 from student_info where 性别 in(select 性别 from student_info where sno=’0001′)
2，查询所有选修课程编号为0002 和0003的学生的学号、姓名和性别
select 学号,姓名,性别 from student_info where 学号 in(select 学号 from grade where 课程编号=’0002′ and 学号 in(select 学号 from grade where 课程编号=’0001′))
3，查询出学号为0001的学生的分数比0002号学生最低分高的课程编号的课程编号和分数
select 课程编号, 分数 from grade where 学号=’0001′ and 分数>(select min(分数) from grade where 学号=’0002′)
→ 多表查询：
1，查询分数在80-90分的学生的学号、姓名、分数
select student_info.学号,student_info.姓名,grade.分数 from student_info,grade where grade.分数 between 80 and 90
2，查询学习了’C语言’课程的学生学号、姓名和分数
select student_info.学号,student_info.姓名,grade.成绩from student_info,grade,curriculum where student_info.学号=grade.学号and grade.课程号=curriculum.课程号and curriculum.课程名=’C语言’
3，查询所有学生的总成绩，要求列出学号、姓名、总成绩，没有选课的学生总成绩为空。
select grade.学号,student_info.姓名,sum(grade.成绩) as 总成绩from student_info,grade where grade.学号=student_info.学号group by grade.学号,student_info.姓名
11，题目、活期存款中，“储户”通过“存取款单”和“储蓄所”发生联系。假定储户包括：账号，姓名，电话，地址，存款额；“储蓄所”包括：储蓄所编号，名称，电话，地址（假定一个储户可以在不同得储蓄所存取款）
1、写出设计以上表格的语句（4分）
2、创建一个触发器TR1完成下面内容：
→ 当向“存取款单”表中插入数据时，如果存取标志＝１则应该更改储户表让存款额加上存取金额，如果存取标志＝0则应该更改储户表让存款额减去存取金额,如果余额不足显示余额不足错误。
CREATE TABLE CREATE TRIGGER tr1 on qukuan after insert
AS
BEGIN
declare @sid nvarchar(50)
declare @type int
declare @qian int
declare @yuer int
select @sid=sid,@type=[type],@m=m from inserted
select @yuer=yuer from cunkuan
if(@type=1)
begin
update cunkuan set yuer=yuer+@qian
end
else
begin
if(@yuer<@qian)
begin
print ‘余额不足’
end
else
begin
update cunkuan set yuer=yuer-@qian
end
end
END
GO
12，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。
→ 要求实现如下处理：
1)，写出自定义函数，要求输入借书卡号能得到该卡号所借书金额的总和
CREATE FUNCTION getSUM
(
@CNO int
)
RETURNS int
AS
BEGIN
declare @sum int
select @sum=sum(price) from BOOKS where bno in (select bno from BORROW where cno=@CNO)
return @sum
END
GO
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。
select CNO,count(BNO) as 借书数量from BORROW group by CNO having count(BNO)>3
3)，查询借阅了”水浒”一书的读者，输出姓名及班级。
select name,class from card where cno in( select cno from borrow where bno in(
select bno from BOOKS where bname=’水浒’))
4)，查询过期未还图书，输出借阅者（卡号）、书号及还书日期。
select CNO,BNO,RDATE from borrow where getdate()>RDATE
5)，查询书名包括”网络”关键词的图书，输出书号、书名、作者。
select bno,bname,author from books where bname like ‘网络%’
6)，查询现有图书中价格最高的图书，输出书名及作者。
select bname,author from books where price in(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。
select cno from borrow where bno in (select bno from books where bname=’计算方法’) and cno not in ( select cno from borrow where bno in(select bno from books where bname=’计算方法习题集’)) order by cno desc
或
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。
update borrow set rdate=dateadd(day,7,rdate) from BORROW where cno in(select cno from card where class=’一班’)
9)，从BOOKS表中删除当前无人借阅的图书记录。
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。(这个不确定对不 90%应该是对的 自己看了下书写的)
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）。
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）。
CREATE VIEW V_VIEW AS select name,bname from books,card,borrow where borrow.cno=card.cno and borrow.bno=books.bno and class=’一班’
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。
select a.cno from borrow a,borrow b where a.cno=b.cno and a.bno in(select bno from books where bname=’计算方法’) and b.bno in(select bno from books where bname=’组合数学’) order by a.cno desc
或
SELECT a.CNO FROM BORROW a,BOOKS b WHERE a.BNO=b.BNO AND b.BNAME IN(‘计算方法’,’组合数学’) GROUP BY a.CNO HAVING COUNT(*)=2 ORDER BY a.CNO DESC
14)，用事务实现如下功能：一个借书卡号借走某书号的书，则该书的库存量减少1，当某书的库存量不够1本的时候，该卡号不能借该书
alter PROCEDURE pro_jieshu
@cno int,
@bno int,
@date datetime
AS
BEGIN
begin tran
declare @quantity int
select @quantity=quantity from books where bno=@bno
insert into borrow values(@cno,@bno,@date)
update books set quantity=@quantity-1 where bno=@bno
if(@quantity>0)
begin
commit tran
end
else
begin
print ‘已无库存’
rollback
end
END
GO
15)，用游标实现将书号为‘A001’的书本的价格提高10元
declare @bno int
declare @bname nvarchar(50)
declare @author nvarchar(50)
declare @price int
declare @quantity int
declare mycursor cursor for select * from books
open mycursor
fetch next from mycursor into @bno,@bname,@author,@price,@quantity
while(@@fetch_status=0)
begin
if(@bno=2)
begin
update books set price=@price+10 where current of mycursor
end
fetch next from mycursor into @bno,@bname,@author,@price,@quantity
end
close mycursor
deallocate mycursor
13，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
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 Student.S# from Student,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 S#,Sname from Student,SC where Student.S#=SC.S# and 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 S# from SC where C# in (select C# from SC where S#=’1002′)
group by S# having count(*)=(select count(*) from SC where S#=’1002′);
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#
);
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sql教程
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成笑笑博客
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′;
一定要复习如何通过CreateTable、Alter Table来创建、修改表结构！
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)
1、select * from t where id=(select max(id) from t)
更简单的就是：select top 1 * from t order by id desc
2、题有问题，应该增加一个主键字段
create table T_Persons(id bigint identity(1,1),name nvarchar(20),age int)
insert into T_Persons(name,age) values(‘小明’,20);
insert into T_Persons(name,age) values(‘小明’,20);
insert into T_Persons(name,age) values(‘小黑’,20);
insert into T_Persons(name,age) values(‘小明’,20);
insert into T_Persons(name,age) values(‘小宏’,20);
–取出name、age相同的id最小的数据
select min(id)
from T_Persons
group by name,age;–按照name和age分组
–把id等于最小值的取出来
select * from T_Persons where id in
(
select min(id)
from T_Persons
group by name,age–按照name和age分组
)
答案：
–把不等于最小id的删掉
delete from T_Persons where id not in
(
select min(id)
from T_Persons
group by name,age–按照name和age分组
)
16，将 TableA 转换为TableB
1）纵表结构 TableA
NameCourseGrade张三语文75张三数学80张三英语90李四语文95李四数学55
2）横表结构 TableB
Name语文数学英语张三758090李四95550
答案：
Select Name,
Sum(case Course when ‘语文’ then Grade else 0 end )as 语文,
Sum(case Course when ‘数学’ then Grade else 0 end )as 数学,
Sum(case Course when ‘英语’ then Grade else 0 end )as 英语,
From TableA
Group by Name
17、
表一：student_info
学号姓名性别出生年月家庭住址备注0001张三男1981-8-9北京null0002汤利利女1988-10-14北京null0003刘小敏男1990-10-23北京null0004杨小虎男1987-7-23北京null
表二：curriculum
课程编号课程名称学分0001计算机基础20002C语言2
表三：Grade
学号课程编号分数00010001800001000290 000390
题目：
→ 查询和学号‘0001’的这位同学性别相同的所有同学的姓名和出生年月？
答案：select 姓名，出生年月 from stduent_info where 性别 in(select 性别 from student_info where sno=’0001′)
→ 查询所有选修课程编号为‘0002’和‘0003’的学生的学号、姓名和性别？
答案：Select 学号，姓名，性别 from student_info where 学号 in(select 学号 from grade where 课程编号=’0002′ and 学号 in(select 学号 from grade where 课程编号=’0001′))
→ 查询出学号为‘0001’ 的学生的分数比‘0002’号学生最低分高的课程编号的课程编号和分数？
答案：Select 课程编号，分数 from grade where 学号=‘0001’ and 分数>(select min(分数) from grade where 学号=’0002′)
→ 查询分数在80-90分的学生的学号、姓名、分数？
答案：Select student_info.学号,student_info.姓名，grade.分数 from student_info,grade where grade.分数 between 80 and 90
→ 查询学习了‘C语言’课程的学生学号、姓名和分数？
答案：Select student_info.学号，student_info.姓名，grade.成绩 from student_info,grade,curriculum where student_info.学号=grade.学号 and grade.课程号=curriculum.课程号 and curriculum.课程名=’C语言’
→ 查询出所有学生的总成绩，要求列出学号、姓名、总成绩，没有选课的学生总成绩为空？
→ Select grade.学号，student_info.姓名，sum(grade.成绩) as 总成绩 from student_info,grade where grade.学号=student_info.学号 group by grade.学号，student_info.姓名
18，
create database MyCompany
go
use MyCompany
go
create table Departments
(
Department_ID int identity(1,1) primary key,
Department_Name nvarchar(50),
)
go
create table Employees
(
Employee_Id int identity(1,1) primary key,
Employee_Name nvarchar(50),
Employee_Job nvarchar(50),
Salary money,
Department_Id int foreign key references Departments(Department_ID)
)
go
insert into Departments values(‘财务部’)
insert into Departments values(‘行政部’)
insert into Departments values(‘开发部’)
insert into Employees values(‘曹操’,’组长’,20000,3)
insert into Employees values(‘刘备’,’经理’,30000,3)
insert into Employees values(‘诸葛亮’,’CEO’,10000,2)
insert into Employees values(‘黄月英’,’职员’,5000,1)
insert into Employees values(‘关羽’,’职员’,8000,3)
insert into Employees values(‘张飞’,’职员’,8000,3)
insert into Employees values(‘赵云’,’职员’,7000,3)
insert into Employees values(‘马谡’,’职员’,4000,3)
相关子查询 和 嵌套子查询
→ 1，对于EMPLOYEES中最低工资小于5000的部门，列出EMPLOYEE_JOB为’职员’的员工的部门号，最低工资，最高工资
答案：select max(SALARY) as 最高工资,min(SALARY) as 最低工资,DEPARTMENT_ID as 部门号 from EMPLOYEES as b where EMPLOYEE_JOB =’职员’ and 5000>(select min(SALARY) from EMPLOYEES as a where a.DEPARTMENT_ID = b.DEPARTMENT_ID) group by b.DEPARTMENT_ID
select * from Employees
→ 2，列出EMPLOYEES表中各部门的部门号，最高工资，最低工资
答案：select MAX(Salary) as 最高工资,MIN(Salary) as 最低工资,Department_Id as 部门编号 from Employees
group by Department_Id
→ 3，列出EMPLOYEES表中各部门EMPLOYEE_JOB为’职员’的员工的最低工资，最高工资
答案：select MAX(Salary) as 最高工资,MIN(Salary) as 最低工资,Department_Id as 部门编号 from Employees
where Employee_Job=’职员’ group by Department_Id
→ 4，对于EMPLOYEES中最低工资小于5000的部门，列出EMPLOYEE_JOB为’职员’的员工的部门号，最低工资，最高工资
答案：select max(SALARY) as 最高工资,min(SALARY) as 最低工资,DEPARTMENT_ID as 部门号 from EMPLOYEES as b
where EMPLOYEE_JOB =’职员’ and 5000>(select min(SALARY) from EMPLOYEES as a where a.DEPARTMENT_ID = b.DEPARTMENT_ID) group by b.DEPARTMENT_ID
select Department_Id,max(salary),min(salary) from dbo.Employees where Department_Id in
(select Department_Id from dbo.Employees where EMPLOYEE_JOB =’职员’ group by Department_Id having min(Salary)<5000)
group by Department_Id,,数据库面试题
→ 5，根据部门号由高而低，工资有低而高列出每个员工的姓名，部门号，工资
答案：select DEPARTMENT_ID as 部门号,EMPLOYEE_NAME as 姓名,SALARY as 工资 from EMPLOYEES
order by DEPARTMENT_ID desc,SALARY asc
→ 6，列出’刘备’所在部门中每个员工的姓名与部门号
答案：select EMPLOYEE_NAME,DEPARTMENT_ID from EMPLOYEES where DEPARTMENT_ID in
(select DEPARTMENT_ID from EMPLOYEES where EMPLOYEE_NAME = ‘刘备’)
→ 7，列出每个员工的姓名，工作，部门号，部门名
答案：elect EMPLOYEE_NAME,EMPLOYEE_JOB,EMPLOYEES.DEPARTMENT_ID,DEPARTMENTS.DEPARTMENT_NAME from EMPLOYEES inner join DEPARTMENTS on EMPLOYEES.DEPARTMENT_ID = DEPARTMENTS.DEPARTMENT_ID
→ 8，列出EMPLOYEES中工作为’职员’的员工的姓名，工作，部门号，部门名
答案：select EMPLOYEE_NAME,EMPLOYEE_JOB,DEPARTMENTS.DEPARTMENT_ID,DEPARTMENT_NAME from EMPLOYEES inner join DEPARTMENTS
on DEPARTMENTS.DEPARTMENT_ID = EMPLOYEES.DEPARTMENT_ID where EMPLOYEE_JOB = ‘职员’
→ 9，对于DEPARTMENTS表中，列出所有部门名，部门号，同时列出各部门工作为’职员’的员工名与工作
答案：select DEPARTMENT_NAME as 部门名,DEPARTMENTS.DEPARTMENT_ID as 部门号,EMPLOYEE_NAME as 员工名,EMPLOYEE_JOB as 工作 from DEPARTMENTS inner join EMPLOYEES
on DEPARTMENTS.DEPARTMENT_ID = EMPLOYEES.DEPARTMENT_ID where EMPLOYEE_JOB = ‘职员’
→ 10，对于工资高于本部门平均水平的员工，列出部门号，姓名，工资，按部门号排序
答案：select a.DEPARTMENT_ID as 部门号,a.EMPLOYEE_NAME as 姓名,a.SALARY as 工资 from EMPLOYEES as a
where a.SALARY>(select avg(SALARY) from EMPLOYEES as b where a.DEPARTMENT_ID = b.DEPARTMENT_ID) order by a.DEPARTMENT_ID
→ 11，对于EMPLOYEES，列出各个部门中平均工资高于本部门平均水平的员工数和部门号，按部门号排序
答案：select count(*) as 员工数,a.DEPARTMENT_ID as 部门号 from EMPLOYEES as a
where a.SALARY>(select avg(SALARY) from EMPLOYEES as b where a.DEPARTMENT_ID = b.DEPARTMENT_ID) group by a.DEPARTMENT_ID order by a.DEPARTMENT_ID
→12，对于EMPLOYEES中工资高于本部门平均水平，人数多与1人的，列出部门号，人数，按部门号排序
答案：select count(*) as 员工数,a.DEPARTMENT_ID as 部门号,avg(SALARY) as 平均工资 from EMPLOYEES as a
where (select count(c.EMPLOYEE_ID) from EMPLOYEES as c where c.DEPARTMENT_ID = a.DEPARTMENT_ID and c.SALARY>(select avg(SALARY) from EMPLOYEES as b where c.DEPARTMENT_ID = b.DEPARTMENT_ID))>1
group by a.DEPARTMENT_ID order by a.DEPARTMENT_ID
→13，对于EMPLOYEES中低于自己工资至少5人的员工，列出其部门号，姓名，工资，以及工资少于自己的人数
答案：select a.DEPARTMENT_ID,a.EMPLOYEE_NAME,a.SALARY,(select count(*) from EMPLOYEES as b
where b.SALARY < a.SALARY) as 人数 from EMPLOYEES as a
where (select count(*) from EMPLOYEES as b where b.SALARY<a.SALARY)<5
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