sql statement oracle database paging query on exercises
- Queries maximum wage of three employee information
select *
from (select * from employees
order by salary desc
)
where rownum<=3;
- Queries wages ranked No. 5 to No. 10 employee information
select * from
(select e.*,rownum r
from employees e
order by salary desc
)
where r between 5 and 10;
- Queries first_name is the third to fifth employee information at the beginning of a capital D
select * from
(select e.*,rownum r
from employees e
where first_name like ‘D%’
)
where r between 3 and 5 ;
Subqueries practice
- Show all wages higher than the 'Allan' (first_name) the names and salaries
select *
from employees
where salary > (select salary from employees where first_name = ‘Allan’);
- Details show the 'Allan' (first_name) employees doing the same work
select *
from employees
where job_id = (select job_id from employees where first_name = ‘Allan’);
- And No. 30 show the same department first_name 'Guy' wages and salaries of employees names and salaries
select *
from employees
where department_id=30 and
salary=(select salary from employees where first_name=‘Guy’);
- Discover all wages higher than the average wage (average wage, including all employees) sales staff ( 'SA_REP') (job_id)
select *
from employees
where job_id=‘SA_REP’ and
salary > (select avg(salary) from employees) ;
Table join queries practice
// query employee information the Employees
// Query department information the Departments
// query the location information of locations
the SELECT * from the Employees
the SELECT * from the Departments
the SELECT * from locations Manger
- Name displays the name and salary of all employees and their departments of
select e.first_name,salary,d.department_name r
from employees e inner join departments d state province
on e.department_id =d.department_id
- Queries in R & D ( 'IT') work of the staff number, name, department, job location
select e.first_name,e.manager_id,department_name r,l.street_address a
from employees e inner join departments d
on e.department_id=d.department_id and d.department_name=‘IT’
inner join locations l
on d.location_id = l.location_id;
-
The name and number of employees of various departments query
the SELECT d.department_name, COUNT (*)
from the Employees E Inner the Join the Departments d
ON e.department_id = d.department_id
Group by d.department_name -
Query basic employee information, additional names of its superiors
. The SELECT * E, e2.first_name
from the Employees E, the WHERE e.manager_id the Employees e2 = e2.employee_id;