【上完1Z0-051的第0章第1章】课程第9次（2016-04-15星期五）：

botang · 发表于 2016-4-16 10:39:33

本帖最后由 botang 于 2016-4-19 08:51 编辑

课程第9次（2016-04-15星期五）：关系型数据库原理，SELECT
【上完1Z0-051的第0章】：关系型数据库原理
【上完1Z0-051的第1章】：SELECT
【1Z0-051】：共2章
【1Z0-052】：共3章

2016-04-15a.sql：

select department_id,rowid, rownum from departments;
select * from departments where rownum <2;
sele * from departments;
select employee_id, salary , salary*(1+nvl(e.commission_pct,0))*12 ,
commission_pct, department_id from employees e;
SELECT department_name || q'[ Department's Manager Id: ]'
|| manager_id
AS "Department and Manager"
FROM departments;
--
select first_name||'AAAA'||last_name "FULL NAME", 9 from employees;
select chr(39) from dual ;
select q'[']' from dual;
SELECT DISTINCT department_id, manager_id
FROM employees;
select * from user_tab_cols tc where tc.TABLE_NAME='EMPLOYEES';
select * from employees;
---
select * from employees where employee_id=100;
select * from employees
where employee_id
not in ( select manager_id from employees where manager_id is not null ) ;

复制代码