问几道问题
1:查询和080603学生相同课程的学生名单怎么写?
我写的这个好像是至少选修了080603学生的学生名单
select distinct stu.* from stu inner join sc using(sid) where
cid in ( select cid from sc where sid='080603') and sid<>'080603' group by sid having count(*)
= (select count(cid) from sc where sid='080603')
2:查询出来各科成绩的前两名和后两名的学生的信息(包括并排名次)
3:查询出来学生的学号,姓名,平均成绩和按照平均成绩排序
select a.*,(select count(*)+1 from t where a.avge>avge) as num from
(select stu.sid,sname,ifnull(sum(score)/count(cid),0) as avge
from stu left join sc
on stu.sid=sc.sid
group by stu.sid)t as a这样写报错"as a "出,什么错误,应该怎么写??
4:网上搜的一个人培训用的几道题,有点看不懂
有两表a和b,前两字段完全相同:(id int,name varchar(10)...),都有下面的数据(当然还有其它字段,这里不列出来了):
id name
----------- ----------
1 a
2 b
3 c 以下的查询语句,你知道它的运行结果吗?:
1.
select * from a left join b on a.id=b.id where a.id=1
2.
select * from a left join b on a.id=b.id and a.id=1
3.
select * from a left join b on a.id=b.id and b.id=1
4.
select * from a left join b on a.id=1
5.
select * from a left join b on a.id=1 where a.id=1
1:查询和080603学生相同课程的学生名单怎么写?
我写的这个好像是至少选修了080603学生的学生名单
select distinct stu.* from stu inner join sc using(sid) where
cid in ( select cid from sc where sid='080603') and sid<>'080603' group by sid having count(*)
= (select count(cid) from sc where sid='080603')
2:查询出来各科成绩的前两名和后两名的学生的信息(包括并排名次)
3:查询出来学生的学号,姓名,平均成绩和按照平均成绩排序
select a.*,(select count(*)+1 from t where a.avge>avge) as num from
(select stu.sid,sname,ifnull(sum(score)/count(cid),0) as avge
from stu left join sc
on stu.sid=sc.sid
group by stu.sid)t as a这样写报错"as a "出,什么错误,应该怎么写??
4:网上搜的一个人培训用的几道题,有点看不懂
有两表a和b,前两字段完全相同:(id int,name varchar(10)...),都有下面的数据(当然还有其它字段,这里不列出来了):
id name
----------- ----------
1 a
2 b
3 c 以下的查询语句,你知道它的运行结果吗?:
1.
select * from a left join b on a.id=b.id where a.id=1
2.
select * from a left join b on a.id=b.id and a.id=1
3.
select * from a left join b on a.id=b.id and b.id=1
4.
select * from a left join b on a.id=1
5.
select * from a left join b on a.id=1 where a.id=1
这是网上的一篇帖子,我还没来的及看,你看一下。
一些高难度的SQL面试题
2008 年 11月 7 日 | 分类:数据库面试题 | 暂无评论 | 601 次阅读| 来源:Net
1.根据如下表的查询结果,那么以下语句的结果是(知识点:not in/not exists+null)
以下的null代表真的null.
SQL> select * from usertable;
USERID USERNAME
———– —————-
1 user1
2 null
3 user3
4 null
5 user5
6 user6
SQL> select * from usergrade;
USERID USERNAME GRADE
———- —————- ———-
1 user1 90
2 null 80
7 user7 80
8 user8 90执行语句:
select count(*) from usergrade where username not in (select username from usertable);
select count(*) from usergrade g where not exists
(select null from usertable t where t.userid=g.userid and t.username=g.username);
结果为:语句1( 0 ) 语句2 ( 3 )
A: 0 B:1 C:2 D:3 E:NULL
2. 在以下的表的显示结果中,以下语句的执行结果是(知识点:in/exists+rownum)
SQL> select * from usertable;
USERID USERNAME
———– —————-
1 user1
2 user2
3 user3
4 user4
5 user5
SQL> select * from usergrade;
USERNAME GRADE
—————- ———-
user9 90
user8 80
user7 80
user2 90
user1 100
user1 80
执行语句
Select count(*) from usertable t1 where username in
(select username from usergrade t2 where rownum <=1);
Select count(*) from usertable t1 where exists
(select ‘x’ from usergrade t2 where t1.username=t2.username and rownum <=1);
以上语句的执行结果是:( ) ( )
A: 0 B: 1 C: 2 D: 3
根据以下的在不同会话与时间点的操作,判断结果是多少,其中时间T1<……
原始表记录为;
select * from emp;
EMPNO DEPTNO SALARY
—– —— ——
100 1 55
101 1 50
select * from dept;
DEPTNO SUM_OF_SALARY
—— ————-
1 105
2
可以看到,现在因为还没有部门2的员工,所以总薪水为null,现在,
有两个不同的用户(会话)在不同的时间点(按照特定的时间顺序)执行了一系列的操作,那么在其中或最后的结果为:
time session 1 session2
———– ——————————- ———————————–
T1 insert into emp
values(102,2,60)
T2 update emp set deptno =2
where empno=100
T3 update dept set sum_of_salary =
(select sum(salary) from emp
where emp.deptno=dept.deptno)
where dept.deptno in(1,2);
T4 update dept set sum_of_salary =
(select sum(salary) from emp
where emp.deptno=dept.deptno)
where dept.deptno in(1,2);
T5 commit;
T6 select sum(salary) from emp group by deptno;
问题一:这里会话2的查询结果为:
T7 commit;
=======到这里为此,所有事务都已完成,所以以下查询与会话已没有关系========
T8 select sum(salary) from emp group by deptno;
问题二:这里查询结果为
T9 select * from dept;
问题三:这里查询的结果为
问题一的结果( ) 问题二的结果是( ) 问题三的结果是( )
A: B:
—————- —————-
1 50 1 50
2 60 2 55
C: D:
—————- —————-
1 50 1 115
2 115 2 50
E: F:
—————- —————-
1 105 1 110
2 60 2 55
有表一的查询结果如下,该表为学生成绩表(知识点:关联更新)
select id,grade from student_grade
ID GRADE
——– ———–
1 50
2 40
3 70
4 80
5 30
6 90
表二为补考成绩表
select id,grade from student_makeup
ID GRADE
——– ———–
1 60
2 80
5 60
现在有一个dba通过如下语句把补考成绩更新到成绩表中,并提交:
update student_grade s set s.grade =
(select t.grade from student_makeup t
where s.id=t.id);
commit;
请问之后查询:
select GRADE from student_grade where id = 3;结果为:
A: 0 B: 70 C: null D: 以上都不对
根据以下的在不同会话与时间点的操作,判断结果是多少,
其中时间T1<……
session1 session2
————————————– —————————————-
T1 select count(*) from t;
–显示结果(1000)条
T2 delete from t where rownum <=100;
T3 begin
delete from t where rownum <=100;
commit;
end;
/
T4 truncate table t;
T5 select count(*) from t;
–这里显示的结果是多少
A: 1000 B: 900 C: 800 D: 0
标签: 数据库, 高难度
我有根据题意设计的表,这是我看小梁的博客50道常见sql查询时的疑问。耽误你的时间了。非常感谢你的帮忙。
mysql> select * from sc2 order by sid,cid;
+--------+--------+-------+
| sid | cid | score |
+--------+--------+-------+
| 080601 | 180101 | 80 |
| 080601 | 180102 | 85 |
| 080601 | 180103 | 82 |
| 080602 | 180101 | 85 |
| 080602 | 180102 | 84 |
| 080602 | 180103 | 86 |
| 080602 | 180104 | 53 |
| 080603 | 180101 | 85 |
| 080603 | 180102 | 82 |
| 080603 | 180103 | 34 |
| 080604 | 180101 | 81 |
| 080604 | 180102 | 83 |
| 080605 | 180101 | 81 |
| 080605 | 180102 | 82 |
+--------+--------+-------+
第一道题,查询和080601选修相同课程的学生信息
select * from sc2 where
cid in ( select cid from sc2 where sid='080601') and sid <>'080601' group by sid having count(*)
= (select count(cid) from sc2 where sid='080601')
080603和080601一样但是上边的写法把080602也选了出来
第二:查询至少有一门课不及格的学生名单这样怎么会有错
select sid,sname from stu where exists (select count(*) >=1 from sc where stu.sid=sc.sid and score<60 );
这样是对的:select sid,sname from stu where 1<=(select count(*) from sc where stu.sid=sid and score<60 );
第三:查询出来180101和180102的前两名和后两名(考虑并排)
比如结果
080601 180101 80
080605 180101 81
080604 180101 81
080603 180101 85
080602 180101 85
080605 180102 82
080603 180102 82
080602 180102 84
080601 180102 85
第四:
可作为安全机制使用。
即使对于没有直接执行存储过程中语句的权限的用户,也可授予他们执行该存储过程的权限。(如果有存储过程的权限而存储过程有增加删除
那么该用户也有增加删除权限了么??)
第四:获得每个人的平均成绩在加上其位序这样写有错
select a.* , (select count(*)+1 from t where a.avge>avge) as num from
( select sid,avg(score) avge from sc group by sid order by avge) t a,应该怎么写???
+--------+--------+-------+
| sid | cid | score |
+--------+--------+-------+
| 080601 | 180101 | 80 |
| 080601 | 180102 | 85 |
| 080601 | 180103 | 82 |
| 080602 | 180101 | 85 |
| 080602 | 180102 | 84 |
| 080602 | 180103 | 86 |
| 080602 | 180104 | 53 |
| 080603 | 180101 | 85 |
| 080603 | 180102 | 82 |
| 080603 | 180103 | 34 |
| 080604 | 180101 | 81 |
| 080604 | 180102 | 83 |
| 080605 | 180101 | 81 |
| 080605 | 180102 | 82 |
+--------+--------+-------+
14 rows in set (0.00 sec)mysql>[/code]第一道题,查询和080601选修相同课程的学生信息
mysql> select distinct sid
-> from sc2 a
-> where not exists (select cid from sc2 where sid=a.sid and cid not in (select cid from sc2 where sid='080601'))
-> and not exists (select cid from sc2 where sid='080601' and cid not in (select cid from sc2 where sid=a.sid));
+--------+
| sid |
+--------+
| 080601 |
| 080603 |
+--------+
2 rows in set (0.00 sec)mysql>
-> from sc2
-> where score<60;
+--------+
| sid |
+--------+
| 080602 |
| 080603 |
+--------+
2 rows in set (0.00 sec)mysql>
-> from sc2 a
-> where (cid='180101' or cid='180102')
-> and 2>(select count(*) from sc2 where cid=a.cid and score>a.score)
-> union all
-> select *
-> from sc2 a
-> where (cid='180101' or cid='180102')
-> and 2>(select count(*) from sc2 where cid=a.cid and score<a.score)
-> order by cid,score
-> ;
+--------+--------+-------+
| sid | cid | score |
+--------+--------+-------+
| 080601 | 180101 | 80 |
| 080604 | 180101 | 81 |
| 080605 | 180101 | 81 |
| 080602 | 180101 | 85 |
| 080603 | 180101 | 85 |
| 080603 | 180102 | 82 |
| 080605 | 180102 | 82 |
| 080602 | 180102 | 84 |
| 080601 | 180102 | 85 |
+--------+--------+-------+
9 rows in set (0.00 sec)mysql>
-> from (
-> select sid,avg(score) as avg_score
-> from sc2
-> group by sid
-> ) a, (
-> select sid,avg(score) as avg_score
-> from sc2
-> group by sid
-> ) b
-> where a.avg_score<=b.avg_score
-> group by a.sid,a.avg_score
-> order by a.avg_score desc;
+--------+-----------+----------+
| sid | avg_score | count(*) |
+--------+-----------+----------+
| 080601 | 82.3333 | 1 |
| 080604 | 82.0000 | 2 |
| 080605 | 81.5000 | 3 |
| 080602 | 77.0000 | 4 |
| 080603 | 67.0000 | 5 |
+--------+-----------+----------+
5 rows in set (0.00 sec)