oracle practice exercises

1. The employee's email query full name, company email unified with "@ zpark.cn" end
   select email || '@ zpark.cn' from employees;
2. According to the old entry date employee information arranged by the New
   select employee_id, first_name, last_name, email , phone_number, hire_date, job_id, salary, commission_pct, manager_id, department_id from employees order by hire_date;
3. Query No. 80 sector wage greater than 7000 employees full name and salary
   select first_name || '' || last_name as name, salary from employees where department_id = 80 and salary> 7000;
4. All last_name employee information inquiry by the more than four-letter
   select * from employees where last_name like ' ____%';

One-way function exercises

1. The column hiredate seen as employee birthdays, birthday this month inquiry staff (visits knowledge: one-way function)
   the SELECT * from the Employees the WHERE the TO_CHAR (hire_date, 'mm') = the TO_CHAR (sysdate, 'mm');
2. Please contact 2002 recruits (study knowledge: one-way function) with three or more ways
   ①select * from the Employees the WHERE the TO_CHAR (hire_date, 'yyyy') = '2002';
   ②select * from the Employees the WHERE hire_date the BETWEEN to_date ( '2002 -1-1 ',' YYYY-mm-dd ') and TO_DATE (' 2002-12-30 ',' YYYY-mm-dd ');
   ③select from the Employees WHERE the hire_date *> = TO_DATE (' 2002-1-1 ',' yyyy-mm-dd ') and hire_date <= to_date (' 2002-12-30 ',' yyyy-mm-dd ');
3. Queries second half of 2002 recruits (study knowledge: one-way function)
   the SELECT * from the Employees to_date the WHERE hire_date the BETWEEN ( '2002-7-1', 'yyyy-mm-dd') and to_date ( '2002-12-30 ',' yyyy-mm-dd ');
4. Print how many days he was born
   select to_date (to_char (sysdate, ' yyyy-mm-dd'), 'yyyy-mm-dd') - to_date ( '1996-3-1', 'yyyy-mm-dd') from dual;

Group Functions practice

1. Employee number (inspection knowledge: group function) seeking 1997 each month recruited a
   select count (*), to_char ( hire_date, 'mm') from employees where to_char (hire_date, 'yyyy') = '2002' group by to_char (hire_date, 'mm') ;
2. Query No. 50 departments, 60 departments, 70 departments of the average salary
   select avg (salary), department_id from employees where department_id in (50, 60, 70) group by department_id;
3. The highest average wage for the wage of 8,000 yuan higher than the sector
   select max (salary), department_id from employees group by department_id having avg (salary)> 8000;
4. Statistics in the number of company managers
   ①select COUNT (COUNT (manager_id)), manager_id from the Employees manager_id the WHERE IS not null Group by manager_id;
   ②select COUNT (DISTINCT manager_id) from the Employees;

Paging query exercise

Queries wages ranked No. 5 to No. 10 employee information
select * from (select employee_id, first_name , rownum as rn from (select * from employees order by salary desc)) where rn> = 5 and rn <= 10;

Subqueries practice

1. Queries salary greater than the average wage of employees in this sector basic information
   select * from (select * from employees group by department_id having salary> avg (salary));
2. Display with 30 departments first_name is 'Guy' wages the same employee names and salaries
   select first_name, salary from employees where salary = (select salary from employees where department_id = 30 and first_name = 'Guy');
3. The number of inquiries each employee wage jobs is greater than the average wage (average wage, including all staff) and staff positions
   select count (*), job_id from employees where salary> (select avg (salary) from employees) group by job_id;

Table join queries practice

1. Display the names of all employees and their departments the names and salaries
   select employees.first_name, department_name, salary from employees left outer join departments on employees.department_id = departments.department_id;
2. Query number of employees working in research and development ( 'IT'), name, department, job location
   select employees.employee_id, employees.first_name, employees.department_id from ( employees left outer join departments on employees.department_id = departments.department_id) left outer join locations on departments.location_id = locations.location_id
3. The name and number of employees of various departments inquiry
   select departments.department_name, count (*) from employees left join departments on employees.department_id = departments.department_id group by departments.department_name;
4. Query basic employee information, additional names of its superior
   select e1.employee_id, e1.first_name, e1.salary, e2.first_name from employees e1 left join employees e2 on e1.manager_id = e2.employee_id
5. Seeking the same entry date (same date) staff (visits knowledge: self-connection) (duplicate)
   the SELECT e1.first_name, e2.first_name from the Employees E1 Inner ON e1.hire_date the Join the Employees e2 = e2.hire_date and E1. ! = e2.first_name FIRST_NAME
   WHERE e1.employee_id <e2.employee_id; - (de-emphasis)
6. Shows the basic salary of each department manager
   select employees.employee_id, salary from departments left outer join employees on departments.manager_id = employees.employee_id