Alexander, that's a classic one,
rewrite your last query as :
SELECT count(employee_id) from employees where employee_id not in
(select manager_id from employees WHERE manager_id IS NOT NULL);
NULLS semantics are sometimes not so obvious.
Στις Monday 03 October 2011 09:33:12 ο/η Alexander Pyhalov έγραψε:
> Hello.
> I was asked a simple question. We have table employees:
> \d employees
> Table "public.employees"
> Column | Type |
> Modifiers
> ----------------+-----------------------------+-----------------------------------------------------------------
> employee_id | integer | not null default
> nextval('employees_employee_id_seq'::regclass)
> first_name | character varying(20) |
> last_name | character varying(25) | not null
> email | character varying(25) | not null
> phone_number | character varying(20) |
> hire_date | timestamp without time zone | not null
> job_id | character varying(10) | not null
> salary | numeric(8,2) |
> commission_pct | numeric(2,2) |
> manager_id | integer |
> department_id | integer |
> Indexes:
> "employees_pkey" PRIMARY KEY, btree (employee_id)
> "emp_email_uk" UNIQUE, btree (email)
> "emp_department_ix" btree (department_id)
> "emp_job_ix" btree (job_id)
> "emp_manager_ix" btree (manager_id)
> "emp_name_ix" btree (last_name, first_name)
> Check constraints:
> "emp_salary_min" CHECK (salary > 0::numeric)
> Foreign-key constraints:
> "employees_department_id_fkey" FOREIGN KEY (department_id)
> REFERENCES departments(department_id)
> "employees_job_id_fkey" FOREIGN KEY (job_id) REFERENCES jobs(job_id)
> "employees_manager_id_fkey" FOREIGN KEY (manager_id) REFERENCES
> employees(employee_id)
> Referenced by:
> TABLE "departments" CONSTRAINT "dept_mgr_fk" FOREIGN KEY
> (manager_id) REFERENCES employees(employee_id)
> TABLE "employees" CONSTRAINT "employees_manager_id_fkey" FOREIGN
> KEY (manager_id) REFERENCES employees(employee_id)
> TABLE "job_history" CONSTRAINT "job_history_employee_id_fkey"
> FOREIGN KEY (employee_id) REFERENCES employees(employee_id)
>
> Now we want to select count of all employees who doesn't have any
> subordinates (query 1):
> SELECT count(employee_id) from employees o where not exists (select 1
> from employees where manager_id=o.employee_id);
> count
> -------
> 89
> (1 row)
>
> We can select count of all managers (query 2):
> SELECT count(employee_id) from employees where employee_id in (select
> manager_id from employees);
> count
> -------
> 18
> (1 row)
>
> But if we reformulate the first query in the same way, answer is
> different (query 3):
> SELECT count(employee_id) from employees where employee_id not in
> (select manager_id from employees) (query 3);
> count
> -------
> 0
> (1 row)
>
> I don't understand why queries 1 and 3 give different results. They
> seems to be the same... Could someone explain the difference?
>
> --
> Best regards,
> Alexander Pyhalov,
> system administrator of Computer Center of Southern Federal University
>
--
Achilleas Mantzios
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