Home > Database > Mysql Tutorial > oracle 查询测试样题

oracle 查询测试样题

WBOY
Release: 2016-06-07 14:51:09
Original
1064 people have browsed it

1.select count(*) from employees where last_name like _A%;key:02.select count(*)from employeeswhere to_char(hire_date,YYYY)=1998;select count(*)from employeeswhere hire_date like %98;key:23select to_char(hire_date,YYYY) from employees;3.se


1.
select count(*) from employees 
where last_name like '_A%';

key:0

2.
select count(*)
from employees
where to_char(hire_date,'YYYY')=1998;

select count(*)
from employees
where hire_date like '%98';

key:23

select to_char(hire_date,'YYYY') from employees;

3.
select job_title, max_salary-min_salary as "SAL_DIEF"
from jobs
order by max_salary-min_salary desc;

select job_title,(max_salary-min_salary) as "SAL_DIEF"
from jobs
order by 2 desc;

19行记录

4.
select count(*)
from employees
where (salary>12000 or salary12000;
--当它判断1000为假时就不判断后面的大于12000了.

select job_id from employees;

select * from employees
where job_id in('ST_MAN','SH_CLERK');
--工作岗位名称要加单引号


5.
select count(*)
from employees
where to_char(hire_date,'YYYY')=1999
and to_char(hire_date,'mm')=02;

key:3

select count(*)
from employees
where to_char(hire_date,'YYYY-MM')='1999-02';

6.

select last_name,salary,
decode(trunc(salary/1500),0,'A',
		   1,'B',
                   2,'C',
		     'D'	
) Grade
from employees
where last_name like'%s';


7.
select d.department_id,d.department_name,l.city
from departments d,locations l
where d.department_id in(10,40,90)
and d.location_id=l.location_id;


8.
select l.city,c.country_name,r.region_name
from locations l,regions r,countries c
where l.location_id=1000 
and l.country_id=c.country_id
and c.region_id=r.region_id;

9.
select  m.last_name "MAN_NAME",nvl(e.last_name,'NO EMPLOYEES') "EMP_NAME"
from employees m,employees e
where m.department_id=100
and m.employee_id=e.manager_id(+);

10行记录

select  m.last_name MAN_NAME,nvl(e.last_name,'NO EMPLOYEES') EMP_NAME
from employees m,employees e
where m.department_id=100
and m.employee_id=e.manager_id(+);

10
select department_id,count(*) NUM
from employees
where salary>8000
group by department_id
;
9行记录

11
select department_id,count(*) NUM
from employees
where salary>5000
group by department_id
having count(*)>3;

3行记录

12
select last_name,salary
from employees
where salary>
(select salary from employees where employee_id=110)
and department_id=100;

2行记录

13
select count(*) NUM
from employees
where commission_pct<all distinct commission_pct from employees where salary>12000
and commission_pct is not null
);
24行记录







</all>
Copy after login


Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template