3． 假設當前資料庫是xscj，要顯示yggl資料庫的employees表中的內容。
select *
from yggl. employees;
select *
from xs as studentinformation;
select distinct kc.課程名,xs_kc.課程號
from kc,xs_kc
where kc.課程號=xs_kc.課程號;
select xs.姓名,成績
from xs join xs_kc on xs.學號=xs_kc.學號
where 課程號 =
'206' and 成績》=
80;
select xs.學號,姓名,課程名,成績
from xs join xs_kc on xs.學號=xs_kc.學號
join kc on xs_kc.課程號 =kc.課程號
where 課程名 =
'計算機基礎' and 成績》=
80;
select a.學號,a.課程號,b.課程號,a.成績
from xs_kc as a join xs_kc as b
on a.成績=b.成績 and a.學號=b.學號 and a.課程號!=b.課程號;
select 課程名
from kc inner join xs_kc
using
(課程號)
;
select xs.
*,課程號,成績
from xs left outer join xs_kc on xs.學號=xs_kc.學號;
select 課程名,xs_kc.
*from xs_kc right join kc on xs_kc.課程號=kc.課程號;
select 學號,姓名,課程號,課程名
from xs cross join kc;
select distinct 課程名,xs_kc.課程號
from kc straight_join xs_kc
on(kc.課程號=xs_kc.課程號)
;
select employees .
*,salary .
*from employees ,salary
where employees.employeeid =salary .employeeid;
select employees .
*,departments .
*from employees ,departments
where employees .departmentid =departments .departmentid;
select departmentname
from departments join employees on departments .departmentid =employees .departmentid
where employees .name =
'王林'
;
select employees .
*from employees join departments on departments .departmentid =employees .departmentid
where departmentname !=
'財務部'
;
select income
from salary left outer join employees on employees.employeeid=salary.employeeid;
select name ,income ,outcome from employees,salary,departments
where employees.employeeid =salary.employeeid
and employees.departmentid =departments.departmentid
and departmentname=
'財務部'
and income >
2000
;
select name,income ,outcome,birthday from employees,salary
where year
(birthday)
<
1966
and departmentid =
(select departmentid from departments where departmentname =
'研發部'
)and employees.employeeid=salary.employeeid;
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