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oracle从入门到精通复习笔记

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描述一个表用 desc employees
过滤重复的部门 select distinct department_id from employees
别名的三种方式: 1、空格 2、加as 3、" "(多个单词组成的别名必须加空格,要么用下划线分开)
条件匹配日期的: where to_char(date,‘yyyy-mm-dd‘)=‘1997-06-07‘
默认格式: where date = ‘7-6月-1997‘
like: where name like ‘%\_%‘ escape ‘\‘ (%:0个或者多个字符,_表示任意一个字符,escape表示转义关键字)
order by salary asc(升序) desc(降序)
多层排序: order by salary asc,name asc (在工资相同的情况下按照名字升序排列)
删除表中字段重复的记录:
delete from job_grades j1
where rowid <> (select min(rowid) from job_grades j2 where j1.grade_level = j2.grade_level);
如果删除表中自然顺序的第15行,下面语句可实现。
(rowid是数据库的一个伪列,建立表的时候数据库会自动为每个表建立ROWID列
用来唯一标识一行记录。rowid是存储每条记录的实际物理地址,对记录的访问是基于ROWID。)
delete from tab where rowid=(
select ii from (select ROWNUM nn,ROWID ii from tab WHERE ROWNUM<=15) WHERE nn=15);

等值连接:
select employee_id,e.department_id,department_name,city
from employees e,departments d,locations l
where e.department_id = d.department_id and d.location_id = l.location_id
或者
select employee_id,e.department_id,department_name,city
from employees e
join departments d on e.department_id = d.department_id
join locations l on d.location_id = l.location_id
等值连接另外方式:
select last_name,department_id,department_name
from employees join departments
--using (department_id) (前提是两表的列名以及列的数据类型要一样)

非等值连接
select employee_id,last_name,salary,grade_level
from employees e,job_grades j
where e.salary between j.lowest_sal and j.highest_sal
左外连接(哪边空就把“+”号放在哪个表上)
select e.employee_id,e.last_name,d.department_name
from employees e,departments d
where e.department_id = d.department_id(+)
左外(右外、满)连接
select employee_id,e.department_id,department_name
from employees e
left outer
--right outer
--full
join departments d on e.department_id = d.department_id
自连接
--查询公司中员工‘Chen’的manager的信息。
select e1.last_name,e2.last_name,e2.email,e2.salary
from employees e1,employees e2
where e1.manager_id = e2.employee_id and lower(e1.last_name) = ‘chen‘
组函数:(注意:只要不是组函数中的列都应该放在group by中,group by中的顺序不做限定,且其中有的列select中可以没有)
select department_id,job_id,avg(salary)
from employees
group by department_id,job_id
order by department_id asc
有组函数的过滤条件用having而不能用where:
select department_id,avg(salary)
from employees
having avg(salary)>6000
group by department_id
order by department_id asc
查询全公司奖金基数的平均值(注意:avg、count默认是不把值为null的计入,此时要想计入的话必须使用nvl()函数):
select avg(nvl(commission_pct,0))
from employees
查询公司在1995-1998年之间,每年雇佣的人数:
select count(*) "total",
count(decode(to_char(hire_date,‘YYYY‘),‘1995‘,1,null)) "1995",
count(decode(to_char(hire_date,‘YYYY‘),‘1996‘,1,null)) "1996",
count(decode(to_char(hire_date,‘YYYY‘),‘1997‘,1,null)) "1997",
count(decode(to_char(hire_date,‘YYYY‘),‘1998‘,1,null)) "1998"
from employees
where to_char(hire_date,‘YYYY‘) in (‘1995‘,‘1996‘,‘1997‘,‘1998‘)
单行子查询:
谁的工资比‘Abel’的高
select last_name,salary
from employees
where salary > (select salary from employees where last_name = ‘Abel‘)
查询员工为Chen的manager的信息
方法一:(使用自连接)
select e2.last_name,e2.manager_id,e2.email,e2.salary
from employees e1,employees e2
where e1.manager_id = e2.employee_id and e1.last_name = ‘Chen‘
方法二:
select employee_id,last_name,email,salary
from employees
where employee_id = (
select manager_id
from employees
where last_name = ‘Chen‘)
返回job_id与141号员工相同,salary比143号员工多的员工姓名、job_id和工资
select employee_id,last_name,email,salary
from employees
where employee_id = (
select manager_id
from employees
where last_name = ‘Chen‘)
返回公司工资最少的员工的last_name,job_id,salary
select last_name,job_id,salary
from employees
where salary = (select min(salary) from employees)
查询最低工资大于50号部门最低工资的部门id和其最低工资
select department_id,min(salary)
from employees
group by department_id
having min(salary)>(select min(salary) from employees where department_id = 50)
多行子查询(in,any,all):
返回其它部门中比job_id为‘IT_PROG’部门任一工资低的员工的员工号、姓名、job_id以及salary
select employee_id,last_name,job_id,salary
from employees
where job_id <>‘IT_PROG‘ and salary < any (select salary from employees where job_id = ‘IT_PROG‘)
order by last_name asc
返回其它部门中比job_id为‘IT_PROG’部门所有(任意)工资低的员工的员工号、姓名、job_id以及salary
select employee_id,last_name,job_id,salary
from employees
where job_id <>‘IT_PROG‘ and salary < any (select salary from employees where job_id = ‘IT_PROG‘)
order by last_name asc

查询平均工资最低的部门信息
select *
from departments
where department_id = (select department_id
from employees
having avg(salary) = (select min(avg(salary))from employees
group by department_id)
group by department_id)
查询平均工资最低的部门信息和该部门的平均工资
select d.* ,(select avg(salary) from employees where department_id = d.department_id)
from departments d
where department_id = (select department_id
from employees
having avg(salary) = (select min(avg(salary))from employees
group by department_id)
group by department_id)
查询平均工资最高的job信息
select *
from jobs
where job_id in (select job_id
from employees
having avg(salary) = (select max(avg(salary))from employees group by job_id)
group by job_id)
查询平均工资高于公司的部门有哪些
select department_id,avg(salary)
from employees
group by department_id
having avg(salary)>(select avg(salary) from employees)
查询公司中所有manager的详细信息
select *
from employees
where employee_id in (select manager_id from employees)
各个部门中,最高工资中最低的那个部门的 最低工资是多少
select min(salary)
from employees
where department_id = (select department_id
from employees
having max(salary)=(select min(max(salary))
from employees
group by department_id)
group by department_id
)
查询平均工资最高的的部门的manager的详细信息:last_name,department_id,email,salary
select last_name,department_id,email,salary
from employees
where employee_id in (select manager_id
from employees
where department_id = (select department_id
from employees
having avg(salary) = (select max(avg(salary))
from employees
group by department_id)
group by department_id))

查询1999年来公司的员工中的最高工资的那个员工的信息
select *
from employees
where salary = (select max(salary)
from employees
where employee_id in (select employee_id
from employees
where to_char(hire_date,‘YYYY‘) = ‘1999‘))
and to_char(hire_date,‘YYYY‘) = ‘1999‘

查看用户定义的表:
select table_name from user_tables;
查询数据库有哪些对象(表、视图、序列、同义词)
select * from user_catalog;
查看用户定义的各种数据库对象
select distinct object_type from user_objects;

创建表的第一种方式:(白手起家)
create table emp1(
id number(10),
name varchar2(20),
salary number(10,2),
hire_date date
)
第二种方式:借助现有的表来创建(数据也会在原有的表中搬过来)
create table emp2
as select employee_id id,last_name name,hire_date,salary
from employees
(where department_id = 80)
(where department_id = 800):不想要任何数据就给它写个没有的部门或者:where 1=2

修改表:
给表追加多一列:
alter table emp1
add (email varchar2(20))
修改表中的列属性:
1、alter table emp1
    modify (id number(15))
2、alter table emp1
  modify (salary number(20,2) default 2000)
删除一个列:
alter table emp1
drop column email
重命名列:
alter table emp1
rename column salary to sal
删除表:
drop table emp5
清空表:
truncate table emp3;(不可以回滚)
delete from emp2;(可以回滚)
重命名表:
rename emp2 to employees2;

表数据的增、删、改:
向表中增加数据
1、insert into emp1 ‘sysdate’或者
    values (1002,‘BB‘,to_date(‘1998-08-08‘,‘yyyy-mm-dd‘),20000)
2、insert into emp1
    values (1002,‘BB‘,to_date(‘1998-08-08‘,‘yyyy-mm-dd‘),null)

3、insert into emp1(employee_id,last_name,hire_date) 注意:这里是只给部分列赋值,有非空约束的必须赋值
     values (1004,‘DD‘,to_date(‘1990-08-08‘,‘yyyy-mm-dd‘))
4、注意:只想赋值一部分列的话,其它必须是允许放空值的列,这里默认salary是NULL(即有非空约束的必须赋值)
insert into emp1(employee_id,last_name,hire_date)
values (1005,‘EE‘,to_date(‘1996-08-08‘,‘yyyy-mm-dd‘))
5、弹窗式的插入数据
insert into emp1(employee_id,last_name,hire_date,salary)
values (&id,‘&last_name‘,‘&hire_date‘,&salary)
基于现有表的记录插入数据
insert into emp1(employee_id,last_name,hire_date,salary)
select employee_id,last_name,hire_date,salary
from employees
where department_id = 80

更新数据:
update emp1
set salary = 22000
where employee_id = 179
更新114员工的工作和工资使其与206号员工相同
1、select employee_id,job_id,salary
from employees1
where employee_id in (114,205)
2、update employees1
set job_id = (
select job_id from employees1 where employee_id = 205
),salary = (
select salary from employees1 where employee_id = 205
)
where employee_id = 114

调整与employee_id为200的员工的job_id相同的
员工的department_id为employee_id为100的员工的department_id
update employees1
set department_id = (select department_id from employees1 where employee_id = 100)
where job_id = (select job_id from employees1 where employee_id = 200)
容易出现的数据完整性的错误如:
update employees
set department_id = 55
where department_id = 100; 问题出现在表中55号部门本来就不存在
从employees表中删除departments部门名称中含有Public字符的部门id
delete from employees1
where department_id = (select department_id from departments where department_name like ‘%Public%‘)

增:
insert into ...
values(...)

insert into ...
select...from...where...
改:
update ...
set ...
where ...
删:
delete from ...
where ...

事务:
commit;
savepoint A;
rollback to savepoint A;
当用户操作表的时候,还没有commit之前,其它用户是不能够对当前的表进行操作的
更改108号员工的信息:使其工资变为所在部门中的最高工资,job变为公司中平均工资最低的 job
update employees
set salary = (select max(salary)
from employees
where department_id = (
select department_id
from employees
where employee_id = 108)
group by department_id),
job_id = (select job_id
from employees
having avg(salary) = (
select min(avg(salary))
from employees
group by job_id )
group by job_id)
where employee_id = 108
删除108号员工所在部门中工资最低的那个员工
delete from employees
where employee_id = (
select employee_id
from employees
where department_id = (select department_id
from employees
where employee_id = 108)
and salary = (select min(salary)
from employees
where department_id =(select department_id
from employees
where employee_id = 108)
)
)
可以优化成:
delete from employees e
where department_id = (select department_id
from employees
where employee_id = 108)
and salary = (select min(salary)
from employees
where department_id = e.department_id
)
使用约束not null和unique创建表:其中在有unique约束中给它赋值多个null,null之间是不冲突的
create table emp3(
--列级约束:
id number(10) constraint emp3_id_uk unique,
name varchar2(20) constraint emp3_name_nn not null,
email varchar2(20),
salary number(10),
--表级约束:
constraint emp3_email_uk unique(email)
)
主键约束:能够唯一的确定一条记录,同样也分表级约束和列级约束,primary key不仅是not null而且unique
create table emp4(
id number(10) constraint emp4_id_pk primary key,
name varchar2(20) constraint emp4_name_nn not null,
email varchar2(20),
salary number(10),
constraint emp4_email_uk unique(email)
)
或者
create table emp4(
id number(10),
name varchar2(20) constraint emp4_name_nn not null,
email varchar2(20),
salary number(10),
constraint emp4_email_uk unique(email),
constraint emp4_id_pk primary key(id)
)
外键约束:(注意:在emp6中插入数据的时候,不能够插入departments表中department_id没有的数据记录。另外,外键引用的列起码要有一个唯一的约束)
create table emp6(
id number(10),
name varchar2(20) constraint emp6_name_nn not null,
email varchar2(20),
salary number(10),
department_id number(10),
constraint emp6_email_uk unique (email),
constraint emp6_id_pk primary key(id),
constraint emp6_dept_id_fk foreign key(department_id) references departments(department_id)
)
向表中插入departments表中存在的department_id(主键)数据
insert into emp6
values(1002,‘AA‘,null,10000,20)

on delete set null:(级联置空:子表中相应的列置空)
on delete cascade:(级联删除:当父表中的列被删除时,子表中相对应的列也被删除)
create table emp7(
id number(10),
name varchar2(20) constraint emp7_name_nn not null,
email varchar2(20),
salary number(10),
department_id number(10),
constraint emp7_email_uk unique (email),
constraint emp7_id_pk primary key(id),
constraint emp7_dept_id_fk foreign key(department_id) references departments(department_id) on delete set null
)
check约束:比如约束工资的范围
create table emp8(
id number(10),
name varchar2(20) constraint emp8_name_nn not null,
email varchar2(20),
salary number(10) constraint emp8_salary check(salary>1500 and salary<30000),
department_id number(10),
constraint emp8_email_uk unique (email),
constraint emp8_id_pk primary key(id),
constraint emp8_dept_id_fk foreign key(department_id) references departments(department_id) on delete set null
)

修改约束:
添加not null约束
alter table emp5
modify (salary number(10,2) not null)
删除约束:
alter table emp5
drop constraint emp5_name_nn
添加unique约束
alter table emp5
add constraint emp5_name_uk unique(name)
无效化约束:
alter table emp3
disable constraint emp3_email_uk
激活约束:
alter table emp3
enable constraint emp3_email_uk
查询约束:(注意:其中条件里的表名要大写)
select constraint_name,constraint_type,search_condition
from user_constraints
where table_name = ‘EMPLOYEES‘
查询定义有约束的列有哪些:
select constraint_name,column_name
from user_cons_columns
where table_name = ‘EMPLOYEES‘

视图:
它实际上是一个虚表,它是依赖于基表的,当视图中的数据更改时,基表中的相应数据也被更改
为什么要使用视图?
答:1、可以控制数据访问 2、简化查询 3、避免重复访问相同的数据

创建视图:
create view empview
as
select employee_id,last_name,salary
from employees
where department_id = 80
基于多张表来创建视图:
create view empview3
as
select employee_id id,last_name name,salary,e.department_name
from employees e,departments d
where e.department_id = d.department_id
修改视图:create or replace
create or replace view empview2
as
select employee_id id,last_name name,department_name
from employees e,departments d
where e.department_id = d.department_id
屏蔽DML操作:with read only (其他用户只能查看,不能增、删、改)
create or replace view empview2
as
select employee_id id,last_name name,department_name
from employees e,departments d
where e.department_id = d.department_id
with read only

简单视图和复杂视图的区别:简单视图没有分组函数,在复杂视图中若使用了组函数创建的,则对它不能使用DML(增、删、改)的操作,因为有些列在基表中原本是不存在的。
创建一个复杂视图:(注意:基表中不存在的列,创建视图时要给它个别名,如下面的平均工资。)
create or replace view empview3
as
select department_name dept_name,avg(salary) avg_sal
from employees e ,departments d
where e.department_id = d.department_id
group by department_name

rownum是一个伪列,跟id有点相关,且有它自己的一个排序。
比如,你想找到表中工资最高的前10位的员工,以下用rownum作为条件是不行的,因为它有自己默认的排列顺序
select rownum,employee_id,last_name,salary
from employees
where rownum <= 10
order by salary desc
要真想用rownum达到查询最高工资的前10位则要如下这样:(注意:rownum只能使用<或<=,而用=,>,>=都不会返回任何数据)
select rownum,employee_id,last_name,salary
from (select employee_id,last_name,salary
from employees
order by salary desc)
where rownum <=10
那么要查询最高工资排列40-50名的数据要怎么办?如下:(此时最外层的rn已经不是伪列了)
select rn,employee_id,last_name,salary
from (select rownum rn,employee_id,last_name,salary
from (select employee_id,last_name,salary
from employees
order by salary desc))
where rn <40 and rn<=50

序列:主要用来提供主键值
创建序列:
create sequence empseq
increment by 10
start with 10
maxvalue 100
cycle
nocache
序列在数据插入表时在主键位置的作用:
先:
create table emp01
as
select employee_id,last_name,salary
from employees
where 1=2
然后:
insert into emp01
values(empseq.nextval,‘BB‘,3300)
修改序列:(能修改增量、最大值、最小值、是否循环以及是否装入内存,如要更改初始值则要通过删除序列重新创建序列,因为改后可能会与之前的数据发生冲突,因为序列是唯一的)
alter sequence empseq
increment by 1
nocycle
序列一以下情况中会出现裂缝:
1、回滚
2、系统出现异常
3、多个表同时使用同一个序列
查询序列:(如果指定了nocache选项,则last_number返回的是序列中下一个有效值)
select sequence_name,min_value,max_value,increment_by,last_number
from user_sequences
删除序列
drop sequence empseq
索引的作用,能够加速 oracle服务器的查询速度,主键和唯一约束中系统默认为它所约束的列创建所引,也可以为非唯一的列手动创建所引,创建好之后系统会自动调用索引,不用你手动编码调用
创建索引:
create index emp01_id_ix
on emp01(employee_id)
删除索引:
drop index emp01_id_ix
什么时候创建索引?
1、列中数据值分布范围很广
2、列经常在where子句或者连接条件中出现
3、表经常被访问而且数据量很大,访问的数据大概占数据总量的2%到4%
什么时候不要创建索引?
1、表很小
2、表经常更新
3、查询的数据大于2%到4%
4、表不经常作为where子句或者连接条件中出现

同义词:
创建同义词
create synonym e for employees
删除同义词
drop synonym e

创建用户和密码:
create user atguigu01
identified atguigu01;
给用户登录数据库的权限,(create table,create sequence,create view,create procedure)
grant create session
to atguigu01 ;
创建用户表空间
alter user atguigu01 quota unlimited(或者5M)
on users
更改用户自己的密码:
alter user atguigu01
identified by atguigu;

角色:使用角色分配给用户权限会更快,角色有什么权限,用户就有什么权限
1、创建角色
create role my_role
2、为角色赋予权限
grant create session,create table,create view
to my_role
3、将角色赋予用户
grant my_role to atguigu02

对象:
1、给atguigu01分配表employees的查询、修改的权限
grant select,update
on scott.employees
to atguigu01
验证上一步的操作:
update scott.employees
set last_name = ‘ABCD‘
where employee_id = 206 (此时改的是scott用户中的表,atguigu只是调用它的表,并没有复制过来)

 

未完,待续......

 

oracle从入门到精通复习笔记,布布扣,bubuko.com

oracle从入门到精通复习笔记

标签:des   style   color   使用   io   数据   for   ar   

原文地址:http://www.cnblogs.com/shellway/p/3912745.html

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