oracle——笔记——简单查询语句

create table emp2 ( empno number(4),
                    ename varchar2(10),
                    job varchar2(9),
                    mgr number(4),
                    hiredate date,
                    sal number(7,2),
                    comm number(7,2),
                    deptno number(2) 
                 );
-----------------------------------------------------------------------------------------------------------

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7369, ‘SMITH‘, ‘CLERK‘, 7902, to_date(‘17-12-1980‘, ‘dd-mm-yyyy‘), 800.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7499, ‘ALLEN‘, ‘SALESMAN‘, 7698, to_date(‘20-02-1981‘, ‘dd-mm-yyyy‘), 1600.00, 300.00, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7521, ‘WARD‘, ‘SALESMAN‘, 7698, to_date(‘22-02-1981‘, ‘dd-mm-yyyy‘), 1250.00, 500.00, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7566, ‘JONES‘, ‘MANAGER‘, 7839, to_date(‘02-04-1981‘, ‘dd-mm-yyyy‘), 2975.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7654, ‘MARTIN‘, ‘SALESMAN‘, 7698, to_date(‘28-09-1981‘, ‘dd-mm-yyyy‘), 1250.00, 1400.00, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7698, ‘BLAKE‘, ‘MANAGER‘, 7839, to_date(‘01-05-1981‘, ‘dd-mm-yyyy‘), 2850.00, null, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7782, ‘CLARK‘, ‘MANAGER‘, 7839, to_date(‘09-06-1981‘, ‘dd-mm-yyyy‘), 2450.00, null, 10);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7788, ‘SCOTT‘, ‘ANALYST‘, 7566, to_date(‘19-04-1987‘, ‘dd-mm-yyyy‘), 3000.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7839, ‘KING‘, ‘PRESIDENT‘, null, to_date(‘17-11-1981‘, ‘dd-mm-yyyy‘), 5000.00, null, 10);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7844, ‘TURNER‘, ‘SALESMAN‘, 7698, to_date(‘08-09-1981‘, ‘dd-mm-yyyy‘), 1500.00, 0.00, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7876, ‘ADAMS‘, ‘CLERK‘, 7788, to_date(‘23-05-1987‘, ‘dd-mm-yyyy‘), 1100.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7900, ‘JAMES‘, ‘CLERK‘, 7698, to_date(‘03-12-1981‘, ‘dd-mm-yyyy‘), 950.00, null, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7902, ‘FORD‘, ‘ANALYST‘, 7566, to_date(‘03-12-1981‘, ‘dd-mm-yyyy‘), 3000.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7934, ‘MILLER‘, ‘CLERK‘, 7782, to_date(‘23-01-1982‘, ‘dd-mm-yyyy‘), 1300.00, null, 10);

=============================================================================================================================

create table dept2 ( deptno number(2),
                     dname varchar2(14),
                     loc varchar2(13)
                   );

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

insert into dept2 (DEPTNO, DNAME, LOC)
values (10, ‘ACCOUNTING‘, ‘NEW YORK‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (20, ‘RESEARCH‘, ‘DALLAS‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (30, ‘SALES‘, ‘CHICAGO‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (40, ‘OPERATIONS‘, ‘BOSTON‘);

============================================================

create table salgrade2 ( grade number,
                         losal number,
                         hisal number
                       );

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

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (1, 700, 1200);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (2, 1201, 1400);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (3, 1401, 2000);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (4, 2001, 3000);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (5, 3001, 9999);


========================================================

select * from emp2;

select * from dept2;

select * from salgrade2;

=====================================
select * from emp2;        --employee2 员工表
select * from dept2;       --department2 部门表
select * from salgrade2;   -- salary grade2 工资等级表

-----------------------------------------------------
emp2

empno    员工编号
ename    员工姓名
job      工作/工种
mgr      manager上级编号
hiredate 入职日期
sal      salary 工资
comm     奖金/津贴
deptno   部门编号
---------------------

dept2

deptno  部门号
dname   部门名称
loc     所在地
---------------------------
salgrade2

grade                   等级
losal lowest salary     最低工资
hisal high salary       最高工资

=============================================


数据查询语句:

select
from
where







查询emp2表中，所有员工的姓名、工资、部门号:

select ename,sal,deptno from emp2;
 





查询emp2表中，工资大于1500 的员工信息:

select * from emp2 where sal > 1500;






 
查询emp2表中，上级是7698 的员工姓名和津贴:

select ename,comm from emp2 where mgr = 7698;







列的别名:

select ename,sal,deptno from emp2;

select ename 姓名,sal 工资,deptno 部门号 from emp2;

select ename as 姓名,sal as 工资,deptno as 部门号 from emp2;

select ename as e,sal as s,deptno as d from emp2;

select ename as "Esc",sal as "alS",deptno as "dDD" from emp2;

select ename as "e%",sal as "s#",deptno as "d()" from emp2;






==============================================================================================================



算术运算 + - * /  （）

支持数值型和日期型(只能加减)数据:






查询每个员工的年薪:

select ename,sal,sal*12 from emp2;





 
给20部门员工，涨薪300之后，工资是多少？:
 
select ename,sal,sal+300 from emp2 where deptno = 20;






 
查询每名员工的 工资奖金和 :
 
select sal,comm,sal+comm from emp2;







运算中如果有空值，那么最后结果为空

--空值替换 nvl()

select sal,comm,nvl(comm,0),sal+nvl(comm,0) from emp2;

select * from emp2;
  
insert into emp2(empno) values(1122);

delete from emp2 where empno = 1122;

select ename,nvl(ename,‘未知‘) from emp2;









连接符 :

select ename,sal from emp2;

select ename||‘的工资是‘||sal||‘，部门是‘||deptno from emp2;
  





 
去重



查询emp中，有多少个部门:

select distinct deptno from emp2; 






查询emp中，有多少种工作

select distinct job from emp2;
  










where
比较运算符

一般的比较运算符 > >= <= < = != <>

特殊的比较运算符

   between……and……
   
   in(……)
   
   like ……  像  模糊查询
   
   is null    空值 
   





select * from emp2 where comm is null;




select * from emp2 where comm is not null;
   
   








     
查询工资在1500 到 3000 的员工信息:

select * from emp2  where sal between 1500 and 3000;  --闭区间
                                      下限     上限 

    
select * from emp2 where sal >= 1500 and sal <= 3000;
   







查询从事CLERK或SALESMAN 的员工信息:

1）字符串
2）关于大小写
    
select * from emp2 where job in(‘CLERK‘,‘SALESMAN‘);
   
select * from emp2 where job = ‘CLERK‘ or job = ‘SALESMAN‘ ;
   
select * from emp2 where sal in(1500,3000);

select * from emp2 where sal = 1500 or sal = 3000;
  









like

select * from emp2 where ename like ‘S%‘;








%
sql里的% 等同于linux的* ，代表零个或多个任意字符
 _ 代表一个任意字符

select * from emp2 where ename like ‘S_ITH‘;
  
select t.*,t.rowid from emp2 t;
  
select emp2.*,rowid from emp2;
  



查询以S%开头的员工信息:

select * from emp2 where ename like ‘S\%%‘ escape ‘\‘;       escape ‘\‘:定义\这个符号后面符号位普通字符串，不是系统里默认的符号。即转义

select * from emp2 where ename like ‘S|%%‘ escape ‘|‘;
  
select * from emp2  where ename like ‘S%\%‘ escape ‘\‘;




escape：

用来转译的，比如数据库中有个表 test字段name 中存了字符‘%‘，那么我查询的时候如果想查询第二位是 ‘%‘的记录就要用到模糊查询，



但是‘%‘是模糊查询的通配符，我不能如下查询：

select * from test where name like ‘_%%‘;




这样系统不会把第二个【%】当做字符，只能转译，转译默认的符号是 【\】

select * from test where name like ‘_\%%‘;




但是可以自定义转义符，这时候就用 escape，比如定义【*】为转义符

select * from test where name like ‘_*%%‘ escape ‘*‘;






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







逻辑运算符 and or not ()

运算优先级：not > and > or ,()优先级最高




查询，20 部门中，从事CLERK 工作的员工:
  
select * from emp2 where deptno = 20 and job = ‘CLERK‘;




查询，30 部门中，奖金为空的员工信息:
  
select * from emp2  where deptno = 30 and comm is null;



  
查询，除10 部门之外，工资大于1500 的员工信息:
  
select * from emp2 where deptno != 10 and sal > 1500;



  
查询，工作是SALESMAN ，或工资不小于3000 的员工:
  
select * from emp2 where job = ‘SALESMAN‘ or sal >= 3000;


  
查询，工作不是SALESMAN ，也不是CLERK 的员工:
  
select * from emp2 where job != ‘SALESMAN‘ and job != ‘CLERK‘;
  
select * from emp2 where job in(‘SALESMAN‘,‘CLERK‘);
  
select * from emp2 where job not in(‘SALESMAN‘,‘CLERK‘);



  
查询，工作是SALESMAN，或，工作是PRESIDENT并且工资大于1500的员工信息:
  
select * from emp2 where job = ‘SALESMAN‘ or job = ‘PRESIDENT‘ and sal > 1500;





查询，工作是SALESMAN或PRESIDNET，并且工资大于1500 的员工信息:

select * from emp2 where (job = ‘SALESMAN‘ or job = ‘PRESIDENT‘) and sal > 1500;

select * from emp2 where job in(‘SALESMAN‘,‘PRESIDENT‘) and sal > 1500;
  











order by 排序 

select
from
where
order by 列名|别名|算术表达式|函数
    
order by 列1,别名,函数

order by 列1,列2 desc;

order by的位置：在整个查询语句的最后多次排序 







select * from emp2 order by sal;  --升序

select * from emp2 order by sal asc;  --升序

select * from emp2 order by sal desc; --降序
 
select ename 姓名,job 工作 from emp2 where deptno = 20 order by 姓名 desc;
 




查询员工信息，结果按照工资奖金和 升序排序:

select emp2.*,sal+nvl(comm,0) from emp2 order by sal+nvl(comm,0);
 



查询员工信息，结果按照部门号排序;如果部门号相同，按照工资降序排序

select * from emp2 order by deptno asc,sal desc;








函数
 
单组函数          一个值对应一个结果 

分组函数/聚合函数 多个值对应一个结果  avg() sum()

单组函数
  --字符函数
  UPPER()   LOWER()   INITCAP()  ---字母大小写
  大写       小写    首字母大写
  
  replace()      substr()    concat()  nvl() nvl2()
    替换           截取       连接
  



select ename,lower(ename),initcap(ename) from emp2;

select t.*,t.rowid from emp2 t;

select * from emp2 where lower(job) = ‘clerk‘;
  
select ename,replace(ename,‘S‘,‘s‘) from emp2;

select ename,substr(ename,2,3) from emp2;
                       起始位,长度
  



查询工作名称以SALES 开头的员工信息:

select * from emp2 where job like ‘SALES%‘;

select * from emp2 where substr(job,1,5) = ‘SALES‘;
  
select ename||‘的工资是‘||sal||deptno from emp2;


select concat(ename,sal) from emp2;

select concat(concat(ename,‘的工资是‘),sal) from emp2;









函数是可以嵌套的

select substr(concat(ename,‘的工资是‘),5,5) from emp2;
  
select comm,nvl(comm,0),nvl2(comm,1000,0) from emp2;
                                 非空替换,空值替换  
                                 
       






                          
数值函数

   round(x[,y])  取整或保留指定小数位，规则：四舍五入
   trunc(x[,y])  取整或保留指定小数位，规则：截断
   mod(x,y) 取模/取余
   
   round(5.72) = 6
   round(5.718,2) = 5.72
   round(04.718,-1) = 0
   
   trunc(5.72) = 5
   trunc(5.718,2) = 5.71
   trunc(05.718,-1) = 0
   
   select round(5.72), round(5.718,2),round(5.718,-1) from dual;
   select trunc(5.718,2),mod(8,4),mod(10,3) from dual;
 

  
dual表
   
作用：语句补全
   
   select * from dual;
   
   select sysdate from dual;
   
   select 12*15,round(1000/23,2) from dual;
   
   select Sys_Context(‘userenv‘,‘db_name‘) from dual;--查看当前数据库 名
   
   select Dbms_Random.random from dual;--获得一个随机数
   
   select Dbms_Random.value(10,20) from dual;--获得一个随机数
   












日期函数

使用insert，新增一条记录：
员工编号：1122
员工工作：SALESMAN
入职日期：2018年7月1号

   insert into 表名(列名) values(列对应的值);
   
   insert into emp(empno,job,hiredate) 
   values(1122,‘SALESMAN‘,‘01-7月-18‘);
   
   insert into emp(empno,job,hiredate) 
   values(1122,‘SALESMAN‘,‘2018-7-1‘);   
   
   select * from emp;
   
   --1)默认日期格式
   
    select * from nls_session_parameters;
    alter session set NLS_DATE_FORMAT=‘YYYY-MM-DD‘;
    alter session set NLS_DATE_FORMAT=‘DD-MON-RR‘;
   
   --2) 日期函数 to_date(日期，格式)
   
   insert into emp(empno,job,hiredate) 
   values(1123,‘SALESMAN‘,to_date(‘10-1-2018‘,‘MM-DD-YYYY‘)); 
   
    --查询入职日期早于1981年9 月30 号的员工信息
    
    select
    from
    where hiredate < to_date()