oracle 基础sql

 

1.数据基础表：

 

 

 

 

 

 

 

 

 

 

 

 

2.简单增删改语句：

insert into t_user(id,username,password,roleid) values(seq_user.nextval,'wind','520',109);

delete from t_user t where t.roleid is null; 

update t_user t set t.username='saturn',t.password='456' where t.id=106; 

 

 3.查询语句：

     1) count|distinct

select count(*) from t_user t; //统计表的记录数

select distinct(t.username) from t_user t; //返回不重复的username

select count(distinct t.username||t.password) from t_user t; 
//统计表不重复username和password的记录数

select t.username,count(t.username) from t_user t group by t.username having count(t.username)>1  //统计重复username的：username及其重复次数

 

 

 

    2）inner join | outer left join | outer right join

t_t1

1	A
2	B
3	C

t_t2

ColumnC	ColumnD
2	x
4	Y

 

a)inner join

Select * from t_t1 a inner join t_t2 b on a.columna=b.columnc;

 

columnA	columnB	columnC	columnD
2	B	2	X

b)left outer join

 

Select * from t_t1 a left outer join t_t2 b on a.columna=b.columnc;

 

columnA	columnB	columnC	columnD
1	A
2	B	2	X
3	C

c)right outer join

 

Select * from t_t1 a right outer join t_t2 b on a.columna=b.columnc;

 

columnA	columnB	columnC	columnD
2	B	2	X
		4	Y

 

d)连接三张表：

Select * from (tableA ainner join tableB b on a.id=b.id) inner join tableC c on c.id=tableA.id 

 

 

3)oracle 分页查询：

select * from (select rownum,t.* from t_user t) where rownum>=1 and rownum<=5;

select * from (select rownum rn,t.* from t_user t order by t.id desc) where rn between 4 and 20; 

 

 

4）多表查询：

select t.* from t_user t,t_role t1 where t.roleid=t1.id and t1.rolename='管理员'; 

 

 

5）oracle时间的处理：

 

默认设置时间类型为：date   采用函数 to_date('2010-09-07','yyyy-mm-dd'),或 to_date('2010-9-5 10:00:00','yyyy-mm-dd HH24:MI:SS')

insert into t_log t values(hibernate_sequence.nextval,104,'新建用户114', to_date('2010-9-5 10:56:47','yyyy-mm-dd HH24:MI:SS'));

select t.* from t_log t where cdate between to_date('2010-09-05','yyyy-mm-dd') and to_date('2010-09-07','yyyy-mm-dd') 

 

 6)top N等问题：

select t.* from t_user t where rownum<=3 order by t.id ; 
//前三最小值

select t.* from t_user t where rownum<=3 order by t.id desc;
//前三最大值

select avg(t.roleid),sum(t.id) from t_user t
//平均值与统计
select t.* from t_log t where t.userid<(select avg(id) from t_user);
//
select t.username,avg(t.roleid) from t_user t group by t.username ;
//统计表中某相同字段username的roleid平均值
select min(t.username),max(t.password) from t_user t 
//最大值与最小值

 

 7)字符函数：

 

select upper(substr(username,1,1))||lower(substr(username,2,length(username)-1)) from t_user 
//将t_user表中的username字段内容的第一个字母大写显示，之后的小写显示

 8)相关函数

PL/SQL Mod,Ceil,floor,round函数
Mod是求余；
Ceil是取靠近值最大的整数；
floor是取靠近值最大的整数；
round是四舍五入；

select mod(10,3) from dual ---1
select mod(4.1,3.1) from dual ---1
select mod(4.9,3.1) from dual ---1.8
select ceil(10/3) from dual ---4
select ceil(3.1) from dual ---4
select floor(3.1) from dual ---3
select floor(3.9) from dual ---3
select round(3.5) from dual ---4
select round(3.1) from dual ---3

 

3.oracle笔试题

题目1：显示各门课程的考场数目和可容纳的考生

各数据表结构如下：
表1：课程表 TC
cid 课程id
cname 课程名
Ctype 考试方式
表2：考场表 TR
Roomid 考场id
address 考场地址
Contain 容纳人数
表3：考场课程对应表 TCR
cid 课程id
roomId 考场id

 

sql：select s.cname,count(s.cname),sum(s.CONTAIN)  from (select * from TC c,TR r,TCR cr where cr.cid=c.cid and cr.roomid=r.roomid) s group by s.cname;

 

写道

题目2：显示考试安排有冲突的考生及考试课程和时间
所谓有冲突，就是指一个考生同一时间（或考试时间有交集）安排了两门及两门以上的考试

表1：课程表 TC
cid 课程id
cname 课程名
Ctype 考试方式
表2：考试表t_exam
eId 考试id
cid 课程id
beginTime 开始时间
endTime 结束时间
表3：考试安排表
studentId 学生id
examId 考试id

 

1) select distinct(t1.eid),t1.btime,t1.etime from t_exam t1, t_exam t2 where ((t1.btime between t2.btime and t2.etime)
  2  or (t1.etime between t2.btime and t2.etime)) and t1.eid!=t2.eid order by t1.eid;

2) select distinct(t1.eid),t1.btime,t1.etime from t_exam t1, t_exam t2 where (t1.btime>t2.btime and t1.btime<t2.etime)
or (t1.etime>t2.btime and t1.etime<t2.etime) or (t1.btime in (select t.btime from t_exam t group by t.btime having count(t.btime)>1)) order by t1.eid;