G-3120: Always use table aliases when your SQL statement involves more than one source.

Major

Maintainability

Reason

It is more human readable to use aliases instead of writing columns with no table information.

Especially when using subqueries the omission of table aliases may end in unexpected behaviors and results.

Also, note that even if you have a single table statement, it will almost always at some point in the future end up getting joined to another table, so you get bonus points if you use table aliases all the time.

Example (bad)

select last_name
      ,first_name
      ,department_name
  from      employee  
       join department using (department_id)
 where extract(month from hire_date) = extract(month from sysdate);

Example (better)

select e.last_name
      ,e.first_name
      ,d.department_name
  from      employee   e
       join department d using (department_id)
 where extract(month from e.hire_date) = extract(month from sysdate);

Example (good)

Using meaningful aliases improves the readability of your code.

select emp.last_name
      ,emp.first_name
      ,dept.department_name
  from      employee   emp
       join department dept using (department_id)
 where extract(month from emp.hire_date) = extract(month from sysdate);

Example Subquery (bad)

If the job table has no employee_id column and employee has one this query will not raise an error but return all rows of the employee table as a subquery is allowed to access columns of all its parent tables - this construct is known as correlated subquery.

select last_name
      ,first_name
  from employee
 where employee_id in (select employee_id
                         from job
                        where job_title like '%manager%');

Example Subquery (good)

If the job table has no employee_id column this query will return an error due to the directive (given by adding the table alias to the column) to read the employee_id column from the job table.

select emp.last_name
      ,emp.first_name
  from employee emp
 where emp.employee_id in (select j.employee_id
                             from job j
                            where j.job_title like '%manager%');