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‎Урок 11.1 Рекурсия/README.md‎

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# Рекурсия
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```sql
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def factorial(x):
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print("x=" + str(x))
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if x == 1:
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return 1
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else:
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return factorial(x-1) * x
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print(factorial(3))
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```
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```sql
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drop table if exists employees;
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```
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```sql
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create table employees (
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employee_id serial primary key,
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full_name varchar not null,
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manager_id int
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);
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```
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```sql
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insert
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into
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employees (
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employee_id,
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full_name,
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manager_id
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)
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values
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(1,
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'michael north',
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null),
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(2,
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'megan berry',
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1),
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(3,
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'sarah berry',
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1),
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(4,
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'zoe black',
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1),
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(5,
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'tim james',
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1),
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(6,
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'bella tucker',
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2),
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(7,
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'ryan metcalfe',
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2),
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(8,
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'max mills',
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2),
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(9,
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'benjamin glover',
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2),
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(10,
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'carolyn henderson',
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3),
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(11,
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'nicola kelly',
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3),
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(12,
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'alexandra climo',
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3),
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(13,
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'dominic king',
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3),
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(14,
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'leonard gray',
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4),
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(15,
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'eric rampling',
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4),
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(16,
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'piers paige',
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7),
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(17,
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'ryan henderson',
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7),
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(18,
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'frank tucker',
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8),
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(19,
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'nathan ferguson',
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8),
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(20,
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'kevin rampling',
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8);
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```
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```sql
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select
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*
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from
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employees;
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```
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```sql
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with recursive subordinates as (
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select
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e.employee_id,
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e.full_name
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from
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employees e
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where
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e.employee_id = 2
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union
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select
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e.employee_id,
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e.full_name
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from
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subordinates s
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join employees e
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on
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s.employee_id = e.manager_id
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)
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select
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*
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from
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subordinates;
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```
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```sql
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with recursive managers as (
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select
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e.employee_id,
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e.full_name,
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e.manager_id
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from
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employees e
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where
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e.employee_id = 18
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union all
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select
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e.employee_id,
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e.full_name,
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e.manager_id
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from
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managers m
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join employees e
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on
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e.employee_id = m.manager_id
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)
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select
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*
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from
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managers;
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```
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```sql
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- 1 этап
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```
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```sql
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select
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e.employee_id,
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e.full_name,
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e.manager_id
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from
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employees e
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where
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e.employee_id = 18;
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```
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- Результат выполнения запроса
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- 18 frank tucker 8
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- общий результат выполнения cte
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- 18 frank tucker 8
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- содержимое managers
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- 18 frank tucker 8
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- 2 этап
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```sql
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select
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e.employee_id,
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e.full_name,
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e.manager_id
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from
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managers m
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join employees e
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on
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e.employee_id = m.manager_id;
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```
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- результат выполнения запроса
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- 8 max mills 2
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- общий результат выполнения cte
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- 18 frank tucker 8
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- 8 max mills 2
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- содержимое managers
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- 8 max mills 2
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- 3 этап
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```sql
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select
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e.employee_id,
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e.full_name,
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e.manager_id
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from
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managers m
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join employees e
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on
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e.employee_id = m.manager_id;
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```
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- результат выполнения запроса
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- 2 megan berry 1
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- общий результат выполнения cte
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- 18 frank tucker 8
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- 8 max mills 2
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- 2 megan berry 1
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- содержимое managers
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- 2 megan berry 1
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- 4 этап
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```sql
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select
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e.employee_id,
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e.full_name,
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e.manager_id
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from
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managers m
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join employees e
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on
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e.employee_id = m.manager_id;
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```
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- результат выполнения запроса
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- 1 michael north
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- общий результат выполнения cte
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- 18 frank tucker 8
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- 8 max mills 2
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- 2 megan berry 1
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- 1 michael north
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- содержимое managers
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- 1 michael north
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- 5 этап
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```sql
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select
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e.employee_id,
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e.full_name,
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e.manager_id
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from
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managers m
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join employees e
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on
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e.employee_id = m.manager_id;
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```
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- результат выполнения запроса
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- общий результат выполнения cte
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- 18 frank tucker 8
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- 8 max mills 2
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- 2 megan berry 1
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- 1 michael north
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```sql
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with recursive factorial as (
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select
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1 as n,
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1 as res
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union
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select
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f.n + 1 as n,
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f.res * (f.n + 1) as res
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from
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factorial f
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where
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f.n + 1 <= 5
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)
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select
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*
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from
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factorial;
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```
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```sql
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with recursive factorial as (
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select
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5 as n,
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5 as res
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union
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select
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f.n - 1 as n,
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f.res * (f.n - 1) as res
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from
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factorial f
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where
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f.n - 1 >= 1
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)
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select
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*
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from
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factorial
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where
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n = 1;
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```

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