sql面试题

1.用一条SQL语句查询出每门课都大于80分的学生姓名

name   kecheng   fenshu
张三     语文       81
张三     数学       75
李四     语文       76
李四     数学       90
王五     语文       81
王五     数学       100
王五     英语       90

A: select distinct name from table where name not in (select distinct name from table where fenshu<=80)

2.学生表如下:
自动编号   学号   姓名课程编号课程名称分数
1        2005001  张三 0001      数学    69
2        2005002  李四 0001      数学    89
3        2005001  张三 0001      数学    69
删除除了自动编号不同,其他都相同的学生冗余信息

A: delete tablename where 自动编号 not in(select min(自动编号) from tablename group by 学号,姓名,课程编号,课程名称,分数)

一个叫department的表，里面只有一个字段name,一共有4条纪录，分别是a,b,c,d,对应四个球对，现在四个球对进行比赛，用一条sql语句显示所有可能的比赛组合.
你先按你自己的想法做一下，看结果有我的这个简单吗？

答：select a.name, b.name
from team a, team b
where a.name < b.name

请用SQL语句实现：从TestDB数据表中查询出所有月份的发生额都比101科目相应月份的发生额高的科目。请注意：TestDB中有很多科目，都有1－12月份的发生额。
AccID：科目代码，Occmonth：发生额月份，DebitOccur：发生额。
数据库名：JcyAudit，数据集：Select * from TestDB

答：select a.*
from TestDB a
,(select Occmonth,max(DebitOccur) Debit101ccur from TestDB where AccID='101' group by Occmonth) b
where a.Occmonth=b.Occmonth and a.DebitOccur>b.Debit101ccur

************************************************************************************

面试题：怎么把这样一个表儿
year  month amount
1991   1     1.1
1991   2     1.2
1991   3     1.3
1991   4     1.4
1992   1     2.1
1992   2     2.2
1992   3     2.3
1992   4     2.4
查成这样一个结果
year m1  m2  m3  m4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4

答案一、
select year,
(select amount from  aaa m where month=1  and m.year=aaa.year) as m1,
(select amount from  aaa m where month=2  and m.year=aaa.year) as m2,
(select amount from  aaa m where month=3  and m.year=aaa.year) as m3,
(select amount from  aaa m where month=4  and m.year=aaa.year) as m4
from aaa  group by year

这个是ORACLE 中做的：
select * from (select name, year b1, lead(year) over
(partition by name order by year) b2, lead(m,2) over(partition by name order by year) b3,rank()over(
partition by name order by year) rk from t) where rk=1;

************************************************************************************

精妙的SQL语句！
精妙SQL语句
作者：不详发文时间：2003.05.29 10:55:05

说明：复制表(只复制结构,源表名：a 新表名：b)

SQL: select * into b from a where 1<>1

说明：拷贝表(拷贝数据,源表名：a 目标表名：b)

SQL: insert into b(a, b, c) select d,e,f from b;

说明：显示文章、提交人和最后回复时间

SQL: select a.title,a.username,b.adddate from table a,(select max(adddate) adddate from table where table.title=a.title) b

说明：外连接查询(表名1：a 表名2：b)

SQL: select a.a, a.b, a.c, b.c, b.d, b.f from a LEFT OUT JOIN b ON a.a = b.c

说明：日程安排提前五分钟提醒

SQL: select * from 日程安排 where datediff('minute',f开始时间,getdate())>5

说明：两张关联表，删除主表中已经在副表中没有的信息

SQL:

delete from info where not exists ( select * from infobz where info.infid=infobz.infid )

说明：--

SQL:

SELECT A.NUM, A.NAME, B.UPD_DATE, B.PREV_UPD_DATE

FROM TABLE1,

(SELECT X.NUM, X.UPD_DATE, Y.UPD_DATE PREV_UPD_DATE

FROM (SELECT NUM, UPD_DATE, INBOUND_QTY, STOCK_ONHAND

FROM TABLE2

WHERE TO_CHAR(UPD_DATE,'YYYY/MM') = TO_CHAR(SYSDATE, 'YYYY/MM')) X,

(SELECT NUM, UPD_DATE, STOCK_ONHAND

FROM TABLE2

WHERE TO_CHAR(UPD_DATE,'YYYY/MM') =

TO_CHAR(TO_DATE(TO_CHAR(SYSDATE, 'YYYY/MM') ¦¦ '/01','YYYY/MM/DD') - 1, 'YYYY/MM') ) Y,

WHERE X.NUM = Y.NUM （+）

AND X.INBOUND_QTY + NVL(Y.STOCK_ONHAND,0) <> X.STOCK_ONHAND ) B

WHERE A.NUM = B.NUM

说明：--

SQL:

select * from studentinfo where not exists(select * from student where studentinfo.id=student.id) and 系名称='"&strdepartmentname&"' and 专业名称='"&strprofessionname&"' order by 性别,生源地,高考总成绩

说明：

从数据库中去一年的各单位电话费统计(电话费定额贺电化肥清单两个表来源）

SQL:

SELECT a.userper, a.tel, a.standfee, TO_CHAR(a.telfeedate, 'yyyy') AS telyear,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '01', a.factration)) AS JAN,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '02', a.factration)) AS FRI,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '03', a.factration)) AS MAR,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '04', a.factration)) AS APR,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '05', a.factration)) AS MAY,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '06', a.factration)) AS JUE,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '07', a.factration)) AS JUL,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '08', a.factration)) AS AGU,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '09', a.factration)) AS SEP,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '10', a.factration)) AS OCT,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '11', a.factration)) AS NOV,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '12', a.factration)) AS DEC

FROM (SELECT a.userper, a.tel, a.standfee, b.telfeedate, b.factration

FROM TELFEESTAND a, TELFEE b

WHERE a.tel = b.telfax) a

GROUP BY a.userper, a.tel, a.standfee, TO_CHAR(a.telfeedate, 'yyyy')

说明：四表联查问题：

SQL: select * from a left inner join b on a.a=b.b right inner join c on a.a=c.c inner join d on a.a=d.d where .....

说明：得到表中最小的未使用的ID号

SQL:

SELECT (CASE WHEN EXISTS(SELECT * FROM Handle b WHERE b.HandleID = 1) THEN MIN(HandleID) + 1 ELSE 1 END) as HandleID

FROM Handle

WHERE NOT HandleID IN (SELECT a.HandleID - 1 FROM Handle a)

*******************************************************************************

有两个表A和B，均有key和value两个字段，如果B的key在A中也有，就把B的value换为A中对应的value
这道题的SQL语句怎么写？

update b set b.value=(select a.value from a where a.key=b.key) where b.id in(select b.id from b,a where b.key=a.key);

***************************************************************************

高级sql面试题

原表:
courseid coursename score
-------------------------------------
1 java 70
2 oracle 90
3 xml 40
4 jsp 30
5 servlet 80
-------------------------------------
为了便于阅读,查询此表后的结果显式如下(及格分数为60):
courseid coursename score mark
---------------------------------------------------
1 java 70 pass
2 oracle 90 pass
3 xml 40 fail
4 jsp 30 fail
5 servlet 80 pass
---------------------------------------------------
写出此查询语句

没有装ＯＲＡＣＬＥ，没试过
select courseid, coursename ,score ,decode（sign(score-60),-1,'fail','pass') as mark from course

完全正确

SQL> desc course_v
Name Null? Type
----------------------------------------- -------- ----------------------------
COURSEID NUMBER
COURSENAME VARCHAR2(10)
SCORE NUMBER

SQL> select * from course_v;

COURSEID COURSENAME SCORE
---------- ---------- ----------
1 java 70
2 oracle 90
3 xml 40
4 jsp 30
5 servlet 80

SQL> select courseid, coursename ,score ,decode(sign(score-60),-1,'fail','pass') as mark from course_v;

COURSEID COURSENAME SCORE MARK
---------- ---------- ---------- ----
1 java 70 pass
2 oracle 90 pass
3 xml 40 fail
4 jsp 30 fail
5 servlet 80 pass

*******************************************************************************

原表:

id proid proname
1 1 M
1 2 F
2 1 N
2 2 G
3 1 B
3 2 A
查询后的表:

id pro1 pro2
1 M F
2 N G
3 B A
写出查询语句

解决方案

sql求解
表a
列 a1 a2
记录 1 a
1 b
2 x
2 y
2 z
用select能选成以下结果吗？
1 ab
2 xyz
使用pl/sql代码实现，但要求你组合后的长度不能超出oracle varchar2长度的限制。
下面是一个例子
create or replace type strings_table is table of varchar2(20);
/
create or replace function merge (pv in strings_table) return varchar2
is
ls varchar2(4000);
begin
for i in 1..pv.count loop
ls := ls || pv(i);
end loop;
return ls;
end;
/
create table t (id number,name varchar2(10));
insert into t values(1,'Joan');
insert into t values(1,'Jack');
insert into t values(1,'Tom');
insert into t values(2,'Rose');
insert into t values(2,'Jenny');

column names format a80;
select t0.id,merge(cast(multiset(select name from t where t.id = t0.id) as strings_table)) names
from (select distinct id from t) t0;

drop type strings_table;
drop function merge;
drop table t;

用sql：

Well if you have a thoretical maximum, which I would assume you would given the legibility of listing hundreds of employees in the way you describe then yes. But the SQL needs to use the LAG function for each employee, hence a hundred emps a hundred LAGs, so kind of bulky.

This example uses a max of 6, and would need more cut n pasting to do more than that.

SQL> select deptno, dname, emps
2 from (
3 select d.deptno, d.dname, rtrim(e.ename ||', '||
4 lead(e.ename,1) over (partition by d.deptno
5 order by e.ename) ||', '||
6 lead(e.ename,2) over (partition by d.deptno
7 order by e.ename) ||', '||
8 lead(e.ename,3) over (partition by d.deptno
9 order by e.ename) ||', '||
10 lead(e.ename,4) over (partition by d.deptno
11 order by e.ename) ||', '||
12 lead(e.ename,5) over (partition by d.deptno
13 order by e.ename),', ') emps,
14 row_number () over (partition by d.deptno
15 order by e.ename) x
16 from emp e, dept d
17 where d.deptno = e.deptno
18 )
19 where x = 1
20 /

DEPTNO DNAME EMPS
------- ----------- ------------------------------------------
10 ACCOUNTING CLARK, KING, MILLER
20 RESEARCH ADAMS, FORD, JONES, ROONEY, SCOTT, SMITH
30 SALES ALLEN, BLAKE, JAMES, MARTIN, TURNER, WARD

also
先create function get_a2;
create or replace function get_a2( tmp_a1 number)
return varchar2
is
Col_a2 varchar2(4000);
begin
Col_a2:='';
for cur in (select a2 from unite_a where a1=tmp_a1)
loop
Col_a2=Col_a2||cur.a2;
end loop;
return Col_a2;
end get_a2;

select distinct a1 ,get_a2(a1) from unite_a
1 ABC
2 EFG
3 KMN

*******************************************************************************

一个SQL 面试题

去年应聘一个职位未果,其间被考了一个看似简单的题,但我没有找到好的大案.
不知各位大虾有无好的解法?

题为:
有两个表, t1, t2,
Table t1:

SELLER | NON_SELLER
----- -----

A B
A C
A D
B A
B C
B D
C A
C B
C D
D A
D B
D C

Table t2:

SELLER | COUPON | BAL
----- --------- ---------
A 9 100
B 9 200
C 9 300
D 9 400
A 9.5 100
B 9.5 20
A 10 80

要求用SELECT 语句列出如下结果:------如A的SUM(BAL)为B,C,D的和,B的SUM(BAL)为A,C,D的和.......
且用的方法不要增加数据库负担,如用临时表等.

NON-SELLER| COUPON | SUM(BAL) ------- --------
A 9 900
B 9 800
C 9 700
D 9 600
A 9.5 20
B 9.5 100
C 9.5 120
D 9.5 120
A 10 0
B 10 80
C 10 80
D 10 80

关于论坛上那个SQL微软面试题

问题：

一百个账户各有100$，某个账户某天如有支出则添加一条新记录，记录其余额。一百天后，请输出每天所有账户的余额信息

这个问题的难点在于每个用户在某天可能有多条纪录，也可能一条纪录也没有（不包括第一天）

返回的记录集是一个100天*100个用户的纪录集

下面是我的思路：

1.创建表并插入测试数据：我们要求username从1-100
CREATE TABLE [dbo].[TABLE2] (
[username] [varchar] (50) NOT NULL , --用户名
[outdate] [datetime] NOT NULL , --日期
[cash] [float] NOT NULL --余额
) ON [PRIMARY

declare @i int
set @i=1
while @i<=100
begin
insert table2 values(convert(varchar(50),@i),'2001-10-1',100)
insert table2 values(convert(varchar(50),@i),'2001-11-1',50)
set @i=@i+1
end
insert table2 values(convert(varchar(50),@i),'2001-10-1',90)

select * from table2 order by outdate,convert(int,username)

2.组合查询语句：
a.我们必须返回一个从第一天开始到100天的纪录集：
如：2001-10-1（这个日期是任意的）到 2002-1-8
由于第一天是任意一天，所以我们需要下面的SQL语句：
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
这里的奥妙在于：
convert(int,username)-1（记得我们指定用户名从1-100 :-))
group by username,min(outdate):第一天就可能每个用户有多个纪录。
返回的结果：
outdate
------------------------------------------------------
2001-10-01 00:00:00.000
.........
2002-01-08 00:00:00.000

b.返回一个所有用户名的纪录集：
select distinct username from table2
返回结果：
username
--------------------------------------------------
1
10
100
......
99

c.返回一个100天记录集和100个用户记录集的笛卡尔集合：
select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)

) as A
CROSS join
(
select distinct username from table2
) as B
order by outdate,convert(int,username)
返回结果100*100条纪录：
outdate username
2001-10-01 00:00:00.000 1
......
2002-01-08 00:00:00.000 100

d.返回当前所有用户在数据库的有的纪录：
select outdate,username,min(cash) as cash from table2
group by outdate,username

order by outdate,convert(int,username)
返回纪录：
outdate username cash
2001-10-01 00:00:00.000 1 90
......
2002-01-08 00:00:00.000 100 50

e.将c中返回的笛卡尔集和d中返回的纪录做left join:
select C.outdate,C.username,
D.cash
from
(
select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
) as A
CROSS join
(
select distinct username from table2
) as B
) as C
left join
(
select outdate,username,min(cash) as cash from table2
group by outdate,username
) as D
on(C.username=D.username and datediff(d,C.outdate,D.outdate)=0)

order by C.outdate,convert(int,C.username)
注意：用户在当天如果没有纪录，cash字段返回NULL，否则cash返回每个用户当天的余额
outdate username cash
2001-10-01 00:00:00.000 1 90
2001-10-01 00:00:00.000 2 100
......
2001-10-02 00:00:00.000 1 90

2001-10-02 00:00:00.000 2 NULL <--注意这里
......

2002-01-08 00:00:00.000 100   50

f.好了，现在我们最后要做的就是，如果cash为NULL，我们要返回小于当前纪录日期的第一个用户余额(由于我们使用order by cash,所以返回top 1纪录即可，使用min应该也可以)，这个余额即为当前的余额：
case isnull(D.cash,0)
when 0 then
(
select top 1 cash from table2 where table2.username=C.username
and datediff(d,C.outdate,table2.outdate)<0
order by table2.cash
)
else D.cash
end as cash

g.最后组合的完整语句就是
select C.outdate,C.username,
case isnull(D.cash,0)
when 0 then
(
select top 1 cash from table2 where table2.username=C.username
and datediff(d,C.outdate,table2.outdate)<0
order by table2.cash
)
else D.cash
end as cash
from
(
select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
) as A
CROSS join
(
select distinct username from table2
) as B
) as C
left join
(
select outdate,username,min(cash) as cash from table2
group by outdate,username
) as D
on(C.username=D.username and datediff(d,C.outdate,D.outdate)=0)

order by C.outdate,convert(int,C.username)

返回结果：
outdate                            username cash
2001-10-01 00:00:00.000    1         90
2001-10-01 00:00:00.000    2        100
......
2002-01-08 00:00:00.000    100                50

***********************************************************************************

取出sql表中第31到40的记录（以自动增长ID为主键）

*从数据表中取出第n条到第m条的记录*/

declare @m int
declare @n int
declare @sql varchar(800)
set @m=40
set @n=31
set @sql='select top '+str(@m-@n+1) + '* from idetail where autoid not in(
select top '+ str(@n-1) + 'autoid from idetail)'
exec(@sql)

select top 10 * from t where id not in (select top 30 id from t order by id ) orde by id

--------------------------------------------------------------------------------

select top 10 * from t where id in (select top 40 id from t order by id) order by id desc

*******************************************************************************

一道面试题,写sql语句

有表a存储二叉树的节点,要用一条sql语句查出所有节点及节点所在的层.
表a
c1 c2 A ----------1
---- ---- / \
A B B C --------2
A C / / \
B D D N E ------3
C E / \ \
D F F K I ---4
E I
D K
C N

所要得到的结果如下

jd cs
----- ----
A 1
B 2
C 2
D 3
N 3
E 3
F 4
K 4
I 4
有高手指导一下,我只能用pl/sql写出来,请教用一条sql语句的写法

SQL> select c2, level + 1 lv
2 from test start
3 with c1 = 'A'
4 connect by c1 = prior c2
5 union
6 select 'A', 1 from dual
7 order by lv;

C2 LV
-- ----------
A 1
B 2
C 2
D 3
E 3
N 3
F 4
I 4
K 4

已选择9行。

题目1

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

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 S#,SN FROM S

WHERE [S#] IN(

SELECT [S#] FROM C,SC

WHERE C.[C#]=SC.[C#]

AND CN='税收基础')

可以用三表连接做:

SELECT S.S#,SN

FROM S,C,SC

WHERE S.S#=SC.S# AND SC.C#=C.C# AND CN='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'

解法二:

SELECT SN,SD

FROM S

WHERE S# IN (SELECT S# FROM SC WHERE 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')

注解:通常问不包括什么的查询用NOT解决.

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([S#]))

注释:右连接后,若没有选全部课程,则存在s#为NULL,故COUNT(*)=COUNT(S#)是有无选全部课程的标志

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

注释:要强调不重复,要在属性前加DISTINCT关键字

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)

注释:此题应该不存在同一学生选同一门课程的情况,DISTINCT关键字应该可以省去.

题目2

问题描述:

已知关系模式:

S (SNO,SNAME) 学生关系。SNO 为学号，SNAME 为姓名

C (CNO,CNAME,CTEACHER) 课程关系。CNO 为课程号，CNAME 为课程名，CTEACHER 为任课教师

SC(SNO,CNO,SCGRADE) 选课关系。SCGRADE 为成绩

1. 找出没有选修过“李明”老师讲授课程的所有学生姓名

--实现代码:

SELECT SNAME FROM S

WHERE NOT EXISTS(

SELECT * FROM SC,C

WHERE SC.CNO=C.CNO

AND CNAME='李明'

AND SC.SNO=S.SNO)

注释:使用EXISTS逐一比较排除选李明老师课的学生.

解法二:

SELECT SNAME

FROM S,SC

WHERE S.SNO=SC.SNO AND CNO NOT IN (SELECT CNO FROM C WHERE CTEACHER='李明')

2. 列出有二门以上(含两门)不及格课程的学生姓名及其平均成绩

--实现代码:

SELECT S.SNO,S.SNAME,AVG_SCGRADE=AVG(SC.SCGRADE)

FROM S,SC,(

SELECT SNO

FROM SC

WHERE SCGRADE<60

GROUP BY SNO

HAVING COUNT(DISTINCT CNO)>=2

)A WHERE S.SNO=A.SNO AND SC.SNO=A.SNO

GROUP BY S.SNO,S.SNAME

注释:此题重点在于先将SC表用WHERE条件筛选出不及格的条目,再用GROUP BY按SNO分组统计出count(SNO)>=2的条目

解法二:

SELECT SNAME,AVG(SCGRADE)

FROM S,SC

WHERE S.SNO=SC.SNO AND SC.SNO NOT IN (SELECT SNO FROM SC WHERE SCGRADE<60 GROUP BY SNO HAVING COUNT(CNO)>=2)

GROUP BY SNAME

3. 列出既学过“1”号课程，又学过“2”号课程的所有学生姓名

--实现代码:

SELECT S.SNO,S.SNAME

FROM S,(

SELECT SC.SNO

FROM SC,C

WHERE SC.CNO=C.CNO

AND C.CNAME IN('1','2')

GROUP BY SNO

HAVING COUNT(DISTINCT CNO)=2

)SC WHERE S.SNO=SC.SNO

注释:上面给出的解法应该是只选过'1'号和'2'号课程的学生姓名,题目要求的解法应该如下:

解法二:

SELECT SNAME

FROM S,SC,C

WHERE S.SNO=SC.SNO AND SC.CNO=C.CNO AND C.CNAME IN('1','2')

4. 列出“1”号课成绩比“2”号同学该门课成绩高的所有学生的学号

--实现代码:

SELECT S.SNO,S.SNAME

FROM S,(

SELECT SC1.SNO

FROM SC SC1,C C1,SC SC2,C C2

WHERE SC1.CNO=C1.CNO AND C1.NAME='1'

AND SC2.CNO=C2.CNO AND C2.NAME='2'

AND SC1.SCGRADE>SC2.SCGRADE

)SC WHERE S.SNO=SC.SNO

注释:此题重点在于SC表的数据要和自己的数据比较,故要采用自连接,同时,需要C表确定SC表的比较条目,上面给出的解答好像少了条件SC1.SNO=SC2.SNO

解法二:

SELECT SNO,SNAME

FROM S

WHERE SNO IN (SELECT SC1.SNO FROM SC SC1,SC SC2,C C1,C C2 WHERE SC1.SNO=SC2.SNO AND SC1.CNO=C1.CNO AND C1.CNAME='1' AND C2.CNO=SC2.CNO AND C2.CNAME='2' AND SC1,SCGRADE>SC2.SCGRADE)

5. 列出“1”号课成绩比“2”号课成绩高的所有学生的学号及其“1”号课和“2”号课的成绩

--实现代码:

SELECT S.SNO,S.SNAME,SC.[1号课成绩],SC.[2号课成绩]

FROM S,(

SELECT SC1.SNO,[1号课成绩]=SC1.SCGRADE,[2号课成绩]=SC2.SCGRADE

FROM SC SC1,C C1,SC SC2,C C2

WHERE SC1.CNO=C1.CNO AND C1.NAME='1'

AND SC2.CNO=C2.CNO AND C2.NAME='2'

AND SC1.SCGRADE>SC2.SCGRADE

)SC WHERE S.SNO=SC.SNO

注释:此题和上题类似,如果采用解法1，将子查询定义为临时表则比较简单,直接在查询项中给出即可,如果是解法二则比较麻烦

解法二:

SELECT SNO,SNAME,SC1.GRADE,SC2.GRADE

FROM S,SC SC1,SC SC2,C C1,C C2

WHERE SNO IN (SELECT SC1.SNO FROM SC SC1,SC SC2,C C1,C C2 WHERE SC1.SNO=SC2.SNO AND SC1.CNO=C1.CNO AND C1.CNAME='1' AND C2.CNO=SC2.CNO AND C2.CNAME='2' AND SC1,SCGRADE>SC2.SCGRADE) AND SNO=SC1 AND SNO=SC2 AND SC1.CNO=C1.CNO AND C1.CNAME='1' AND SC2.CNO=C2.CNO AND C2.CNAME='2'

题目三

说明：有三个表，项目表、合同表、付款表

--下面是建立表的语句
create table 项目(项目编号 int,项目名称 varchar(50))
insert into 项目
select 1, '项目1' from dual
union all
select 2, '项目2' from dual
union all
select 3, '项目3' from dual;

create table 合同(合同编号 int,项目编号 int,合同金额 number(7,3))

insert into 合同
select 1,1,1000 from dual
union all
select 2,1,1500 from dual
union all
select 3,2,2000 from dual;

create table 付款(付款编号 int,合同编号 int,付款金额 number(7,3))

insert into 付款
select 1,1,100 from dual
union all
select 2,2,200 from dual
union all
select 3,2,800 from dual;

一个项目可能会有签署多个合同，每个合同会分几次付款，

问题(一)
设计一个查询，要求返回结果如下：
项目编号        项目名称             项目所有合同的金额
----------- ----------------------- ------------------------
1           项目1                    2500
2           项目2                    2000
3           项目3                    NULL

解答：
SELECT L.项目编号,MAX(L.项目名称),SUM(R.合同金额) AS 合同金额
FROM 项目 L LEFT OUTER JOIN 合同 R
ON L.项目编号=R.项目编号
GROUP BY L.项目编号

注释:出现NULL值,应该采用外连接,连接后按项目编号分组对付款金额求和,上面给出解法未排序以及聚合函数使用错误,应该用SUM而非MAX

解法二:

SELECT X.项目编号,项目名称,SUM(付款金额)

FROM 项目 X LEFT JOIN 合同 H ON X.项目编号=H.项目编号

GROUP BY X.项目编号

ORDER BY X.项目编号 ASC

问题(二)
设计一个查询，要求返回结果如下：
项目编号        项目所有合同已付款金额
----------- -----------------------------
1           1100
2           NULL
-----------------------------------------

解答：
SELECT T2.项目编号,SUM(T3.付款金额) AS 项目所有合同已付款金额
FROM 合同 T2 LEFT OUTER JOIN 付款 T3
ON T2.合同编号=T3.合同编号
GROUP BY T2.项目编号

解析:同上题,有NULL,先外连接再分组.

解法二:

SELECT H.项目编号,SUM(F.付款金额)

FROM 合同 H LEFT JOIN 付款 F ON H.合同编号=F.合同编号

GROUP BY H.项目编号

ORDER BY H.项目编号 ASC

问题(三)
设计一个查询，要求返回结果如下：
项目编号    项目名称        项目所有合同已付款金额
----------- ----------------------------------------
1           项目1   1100
2           项目2   NULL
----------------------------------------------------

解答：
SELECT T1.项目编号
,MAX(T1.项目名称)
,SUM(T3.付款金额) AS 项目所有合同已付款金额
FROM 项目 T1 JOIN 合同 T2
ON T1.项目编号=T2.项目编号
LEFT OUTER JOIN 付款 T3
ON T2.合同编号=T3.合同编号
GROUP BY T1.项目编号

解析:在上题解法二的基础上建立临时表,再连接项目表即可

SELECT X.项目编号,X.项目名称,HF.合同已付款

FROM 项目 X,(

SELECT H.项目编号编号,SUM(F.付款金额) 合同已付款

FROM 合同 H LEFT JOIN 付款 F ON H.合同编号=F.合同编号

GROUP BY H.项目编号

) HF

WHERE X.项目编号=HF.编号

ORDER BY X.项目编号

问题(四)
请您设计一个查询语句，检索的格式如下

------------------------------------------------------------------------
项目编号    项目名称    项目所有合同的金额    项目所有合同已付款金额
1           项目1        2500.000              1100.000
2           项目2        2000.000              NULL
------------------------------------------------------------------------

解答：
select L.项目编号,L.项目名称,L.项目所有合同的金额,R.项目所有合同已付款金额
from
(select A.项目编号,A.项目名称,sum(B.合同金额) 项目所有合同的金额
   from 项目 A LEFT JOIN 合同 B ON A.项目编号=B.项目编号
     group by A.项目编号,A.项目名称) L
JOIN
(select B.项目编号, sum(付款金额) 项目所有合同已付款金额
   from 合同 B LEFT JOIN 付款 C ON B.合同编号=C.合同编号
     group by B.项目编号) R
ON L.项目编号=R.项目编号

解析:在前两题的基础上进行内联

附:

SQL查询的基本结构

SELECT [列名1],[列名2]

FROM [表名]

WHERE [条件]

GROUP BY [列名] HAVING [分组条件]

ORDER BY [列名] ASC/DESC