oracle查询测试样题

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 salary<1000)
and job_id !='ST_MAN' and job_id!='SH_CLERK' ;


number:6

================
select count(*)
from employees
where salary not between 1000 and 12000
and job_id not in('ST_MAN','SH_CLERK') ;

select salary
from employees
where salary<1000 
and salary>12000;
--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_pct12000
and commission_pct is not null
);
24行记录