Oracle Class notes-day 12th

Grouping Functions , multi-line functions

Summary of non-null data

Only Lieri the null value first.

Grouping is not required

Sql> Select COUNT (*), sum (Salary), AVG (Salary), min (Salary), Max (salary) from employees;

sql> CREATE TABLE t1 (x int);

sql> INSERT INTO T1 values (null);

sql> INSERT INTO T1 values (1);

Sql> commit;

Sql> Select COUNT (*) from T1;

Sql> select COUNT (x) from T1;

Sql> select Max (x) from T1;

sql> Select min (x) from T1;

sql> select SUM (x) from T1;

Sql> Select AVG (x) from T1;

Sql> Select AVG (Salary), AVG (NVL (commission_pct, 0)) from employees;

Sql> Select count (distinct department_id) from employees; Remove duplicate values

GROUP BY group:

Sql> Select department_id, avg (salary) from Employees group by DEPARTMENT_ID;

Multi-column grouping:

Sql> Select department_id, job_id, Max (salary) from Employees group by DEPARTMENT_ID, job_id;

Sql> Select department_id, job_id, Max (Salary), last_name from Employees GROUP by DEPARTMENT_ID, job_id; Error syntax

Practice:

Number of different positions in the company

Select (distinct job_id) from employees;

Calculate the number of people in each department

Select Department_id,count (employees_id)

From Employees

Where department_id is not null

Group by department_id

GROUP by year for the sum of employees ' wages

Select Extract(year fromhire_date), avg (Salary)

From Employees

Group by extract (year from hire_date)

Having statement:

Sql> Select department_id, avg (salary) from employees where AVG (salary) >=5000 GROUP by department_id; Error statement

Sql> Select department_id, avg (salary) from Employees GROUP by DEPARTMENT_ID have avg (salary) >=5000;

Practice:

According to the department to find all the ordinary employees have the average wage, the department's average salary of less than 5000 does not show, the final results in descending order of average wages.

Select department_id, Avg (Salary) Avg_sal

From Employees

where job_id not as '%\_mgr ' escape ' \ ' and department_id is not NULL

GROUP BY department_id

Having avg (Salary) >=5000

ORDER BY avg_sal Desc;

Multi-table connection No

Emp:dept:

Empno ename deptno deptno dname

ABC Ten Sales

101 Def Ten Market

102 XYZ It

103 OPQ NULL

For EMP in 100.. 103

For dept in 10.. 30

Emp.deptno=dept.deptno

ABC Ten Sales

101 def Ten Sales

102 XYZ

Get the following information to prepare for the job:

Employees

Total Employees: 107

Sql> Select COUNT (*) from employees;

Number of employees with Department: 106

Sql> Select COUNT (*) from employees where department_id are not null;

Sql> Select COUNT (department_id) from employees;

Number of employees without department: 1

Sql> Select COUNT (*) from employees where department_id is null;

Departments:

Number of departments: 27

Sql> Select COUNT (*) from departments;

Number of departments with employees: 11

Sql> Select COUNT (distinct department_id) from employees;

Number of departments with no employees: 16

Sql> Select COUNT (*) from departments where department_id not in (select department_id from Employees where Department_ ID is not NULL);

For dept in 1..27

For EMP in 1..107

Dept.deptid does not appear in the EMP table

Select COUNT (*)

From Employees e, departments D

where e.department_id (+) =d.department_id

and e.employee_id is null;

Select COUNT (*)

From Departments D

Where NOT EXISTS

(select 1 from Employees where department_id=d.department_id);

Select (select COUNT (*) from Departments) – (select COUNT (distinct department_id) from employees) from dual;

Internal connections: 106 (106, 11)

Select E.last_name, D.department_name

From Employees e, departments D

where e.department_id=d.department_id;

Select E.last_name, D.department_name

From Employees e joins departments d on E.DEPARTMENT_ID=D.DEPARTMENT_ID;

Left Outer connection: 107 (106+1)

Select E.last_name, D.department_name

From Employees e, departments D

where e.department_id=d.department_id (+);

Select E.last_name, D.department_name

From Departments D, employees E

where e.department_id=d.department_id (+);

Select E.last_name, D.department_name

From employees e LEFT OUTER JOIN departments D

On e.department_id=d.department_id;

Right outer connection: 122 (106+16)

Select E.last_name, D.department_name

From Employees e, departments D

where e.department_id (+) =d.department_id;

Select E.last_name, D.department_name

From Employees e right OUTER join departments D

On e.department_id=d.department_id;

Full Outer connection: 123 (106+1+16)

Select E.last_name, D.department_name

From Employees e full OUTER JOIN departments D

On e.department_id=d.department_id;

Extensions for multi-table joins:

N-Sheet Connection:

Select E.last_name, D.department_name, l.city

From Employees e, Departments D, Locations L

where e.department_id=d.department_id

and d.location_id=l.location_id;

Select E.last_name, D.department_name, l.city

From Employees e join Departments D on e.department_id=d.department_id

Join locations L on d.locationy_id=l.location_id;

Select E.last_name, D.department_name, l.city

From Employees e, Departments D, Locations L

where e.department_id=d.department_id (+)

and d.location_id=l.location_id (+);

Select E.last_name, D.department_name, l.city

From employees e left outer joins departments D on E.DEPARTMENT_ID=D.DEPARTMENT_ID

Left outer joins locations L on d.location_id=l.location_id;

Oracle Class notes-day 12th