reboot rep

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ada-dmitry
2024-02-15 15:29:30 +03:00
parent 2360cb388f
commit 4623603411
25 changed files with 505 additions and 99 deletions
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/* Вариант 1 || Работа 6 || Антипенко, Дробышевский */
-- Task 1 --
/* a) Напишите запрос, используя конструкцию WITH, выбирающий
рекурсивно сотрудника с идентификатором 1 и все его подчинённых,
как прямых, так и подчинённых более низкого ранга*.
b) Напишите программу на языке PL/SQL, печатающую на экран фамилию
и имя сотрудника с идентификатором 1 и всех его подчинённых, как
прямых, так и подчинённых более низкого ранга. */
-- Пункт а)
WITH RECURSIVE tmp AS (
SELECT id, first_name, last_name, manager_id, 1 AS level
FROM bd6_employees
WHERE id = 1
UNION all
SELECT e.id, e.first_name, e.last_name, e.manager_id, t.level + 1
FROM bd6_employees e
JOIN tmp t ON e.manager_id = t.id
)
SELECT *
FROM tmp
;
-- Пункт б)
CREATE OR REPLACE PROCEDURE depa(integer) AS $$
DECLARE
d_attrs RECORD;
i integer;
l integer;
BEGIN
FOR d_attrs IN
(
SELECT id,last_name, first_name, manager_id
FROM bd6_employees
ORDER BY manager_id
)
LOOP
IF d_attrs.manager_id=$1 THEN
RAISE INFO ' % % % % ', d_attrs.first_name, d_attrs.last_name, d_attrs.id, d_attrs.manager_id;
l:= d_attrs.id;
CALL depa(l);
END IF;
END LOOP;
END $$ LANGUAGE plpgsql;
CREATE OR REPLACE PROCEDURE dep(integer) AS $$
DECLARE
d_attrs RECORD;
BEGIN
FOR d_attrs IN
(
SELECT
id,
last_name,
first_name,
manager_id
FROM bd6_employees
ORDER BY manager_id
)
LOOP
IF d_attrs.id=$1 THEN
RAISE INFO ' % % % % ', d_attrs.first_name, d_attrs.last_name, d_attrs.id, d_attrs.manager_id;
END IF;
END LOOP;
CALL depa($1);
END $$ LANGUAGE plpgsql;
CALL dep(1);
-- Task 2 --
/* Напишите программу на языке PL/SQL, выбирающую строки из таблицы
employees в порядке возрастания заработной платы (salary) и печатающие на
экран следующие данные: фамилия, имя, модифицированная заработная
плата. Модифицированная заработная плата получается следующим
образом: у первого по порядку сотрудника она округляется до сотен в
меньшую сторону, а у всех последующих сотрудников она сначала
увеличивается на остаток от округления, полученный от предыдущего
сотрудника, а затем округляется до сотен в меньшую сторону. */
UPDATE bd6_employees SET salary_in_euro = 299 WHERE id = 3;
CREATE OR REPLACE PROCEDURE print_employees() AS $$
DECLARE
current_salary INTEGER;
m_salary INTEGER;
mod_salary INTEGER := 0;
employee_rec RECORD;
BEGIN
FOR employee_rec IN
(
SELECT last_name, first_name, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro
)
LOOP
current_salary := employee_rec.salary_in_euro;
m_salary := ((current_salary + mod_salary) / 100) * 100;
RAISE NOTICE 'Employee: % % || Modified Salary: %', employee_rec.last_name, employee_rec.first_name, m_salary;
mod_salary := mod(current_salary + mod_salary,100);
END LOOP;
END $$ LANGUAGE plpgsql;
CALL print_employees();
SELECT *
FROM bd6_employees
ORDER BY salary_in_euro;
-- Task 3 --
/* Напишите программу на языке PL/SQL, удаляющую 10 сотрудников с
самой маленькой заработной платой. При этом их заработная плата должна
добавиться 10 сотрудникам с самой большой заработной платой. Причём
самая маленькая заработная плата должна добавиться к человеку с самой
большой заработной платой. 2-я с конца заработная плата должна добавиться
к человеку со второй по размеру заработной платой и т.д. */
CREATE OR REPLACE FUNCTION rearrange_salaries() RETURNS VOID AS $$
DECLARE
salaries RECORD;
small NUMERIC[];
larg NUMERIC[];
const CONSTANT INTEGER = 3;
BEGIN
FOR salaries IN
SELECT id, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro
LIMIT const
LOOP
small = array_append(small, (SELECT salary_in_euro FROM bd6_employees WHERE id = salaries.id));
DELETE FROM bd6_employees
WHERE id = salaries.id;
END LOOP;
FOR salaries IN
SELECT id, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro DESC
LIMIT const
LOOP
larg = array_append(larg, (SELECT id FROM bd6_employees WHERE id = salaries.id));
END LOOP;
FOR i IN 1..const
LOOP
UPDATE bd6_employees
SET salary_in_euro = salary_in_euro + small[i]
WHERE id = larg[i];
END LOOP;
END;
$$ LANGUAGE plpgsql;
SELECT rearrange_salaries();
-- Task 4 --
/* Создайте таблицу spiral с 5 полями f1, f2, f3, f4, f5 целые числа.
Напишите программу на языке PL/SQL, заполняющую данную таблицу 1000
строк по следующему принципу:
1 2 3 4 5
10 9 8 7 6
11 12 13 14 15
20 19 18 17 16
21 22 23 24 25 */
CREATE OR REPLACE PROCEDURE f() AS $$
DECLARE d_attrs RECORD;
BEGIN
FOR d_attrs IN WITH RECURSIVE tmp AS (
SELECT 1+i*5 AS f1, 2+i*5 AS f2, 3+i*5 AS f3, 4+i*5 AS f4,5+i*5 AS f5
FROM generate_series(0,199) i)
SELECT * FROM tmp
LOOP
case
when ( mod(d_attrs.f1,2)!=0)
then RAISE INFO ' % % % % %', d_attrs.f1, d_attrs.f2, d_attrs.f3, d_attrs.f4, d_attrs.f5;
when ( mod(d_attrs.f1,2)=0)
then RAISE INFO ' % % % % %', d_attrs.f5, d_attrs.f4, d_attrs.f3, d_attrs.f2, d_attrs.f1;
end case;
END LOOP;
END $$ LANGUAGE plpgsql;
call f();
-- Task 7 from lab4 --
CREATE table staff(
id integer primary key,
name varchar(64) NOT NULL,
department varchar(64) NOT NULL
);
CREATE SEQUENCE staff_id_seq
START WITH 12
CYCLE
INCREMENT BY 8
CACHE 100;
INSERT INTO staff VALUES
(nextval('staff_id_seq'), 'Ivan Makarenko', 'Director');
INSERT INTO staff
(SELECT nextval('staff_id_seq'), last_name, department_name
FROM employees e JOIN departments d ON e.department_id = d.department_id);
UPDATE staff SET department = 'Innovations department'
WHERE id < 40;
DELETE FROM staff
WHERE name ILIKE('%K%');
CREATE VIEW dep_staff_counts AS
SELECT department, COUNT(*) AS ecount
FROM staff
GROUP BY department;
SELECT * FROM dep_staff_counts;
-- CREATE OR REPLACE PROCEDURE department_names(integer) AS $$
-- DECLARE
-- d_attrs RECORD;
-- i INTEGER;
-- l INTEGER;
-- BEGIN
-- FOR d_attrs IN
-- (
-- SELECT id,last_name, first_name, manager_id
-- FROM bd6_employees order by manager_id
-- )
-- LOOP
-- IF d_attrs.manager_id=$1 THEN
-- RAISE INFO ' % % % % ', d_attrs.first_name, d_attrs.last_name, d_attrs.id, d_attrs.manager_id;
-- l:= d_attrs.id;
-- CALL department_names(l);
-- END IF;
-- END LOOP;
-- END $$ LANGUAGE plpgsql;
-- CREATE OR REPLACE PROCEDURE department_names2(integer) AS $$
-- DECLARE d_attrs RECORD;
-- BEGIN
-- FOR d_attrs IN
-- (
-- SELECT id,last_name, first_name, manager_id
-- FROM bd6_employees
-- order by manager_id
-- )
-- LOOP
-- if d_attrs.id=$1 then
-- RAISE INFO ' % % % % ', d_attrs.first_name, d_attrs.last_name, d_attrs.id, d_attrs.manager_id;
-- end if;
-- END LOOP;
-- CALL department_names($1);
-- END $$ LANGUAGE plpgsql;
-- CALL department_names2(4);
/* Вариант 1 || Работа 6 || Антипенко, Дробышевский */
-- Task 1 --
-- Пункт а)
WITH RECURSIVE tmp AS (
SELECT id, first_name, last_name, manager_id, 1 AS level
FROM bd6_employees
WHERE id = 1
UNION all
SELECT e.id, e.first_name, e.last_name, e.manager_id, t.level + 1
FROM bd6_employees e
JOIN tmp t ON e.manager_id = t.id
)
SELECT *
FROM tmp
;
-- Пункт б)
CREATE OR REPLACE PROCEDURE depa(integer) AS $$
DECLARE
d_attrs RECORD;
i integer;
l integer;
BEGIN
FOR d_attrs IN
(
SELECT id,last_name, first_name, manager_id
FROM bd6_employees
ORDER BY manager_id
)
LOOP
IF d_attrs.manager_id=$1 THEN
RAISE INFO ' % % % % ', d_attrs.first_name, d_attrs.last_name, d_attrs.id, d_attrs.manager_id;
l:= d_attrs.id;
CALL depa(l);
END IF;
END LOOP;
END $$ LANGUAGE plpgsql;
CREATE OR REPLACE PROCEDURE dep(integer) AS $$
DECLARE
d_attrs RECORD;
BEGIN
FOR d_attrs IN
(
SELECT
id,
last_name,
first_name,
manager_id
FROM bd6_employees
ORDER BY manager_id
)
LOOP
IF d_attrs.id=$1 THEN
RAISE INFO ' % % % % ', d_attrs.first_name, d_attrs.last_name, d_attrs.id, d_attrs.manager_id;
END IF;
END LOOP;
CALL depa($1);
END $$ LANGUAGE plpgsql;
CALL dep(1);
-- Task 2 --
UPDATE bd6_employees SET salary_in_euro = 299 WHERE id = 3;
CREATE OR REPLACE PROCEDURE print_employees() AS $$
DECLARE
current_salary INTEGER;
m_salary INTEGER;
mod_salary INTEGER := 0;
employee_rec RECORD;
i INTEGER = 0;
cnt INTEGER;
BEGIN
SELECT COUNT(*) INTO cnt FROM bd6_employees;
FOR employee_rec IN
(
SELECT last_name, first_name, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro
)
LOOP
current_salary := employee_rec.salary_in_euro;
IF cnt - 1 = i THEN
m_salary := ((current_salary) / 100) * 100 + mod_salary;
ELSE
m_salary := ((current_salary) / 100) * 100;
END IF;
RAISE NOTICE 'Employee: % % || Modified Salary: %', employee_rec.last_name, employee_rec.first_name, m_salary;
mod_salary := mod_salary + mod(current_salary, 100);
i := i+1;
END LOOP;
END $$ LANGUAGE plpgsql;
CALL print_employees();
SELECT *
FROM bd6_employees
ORDER BY salary_in_euro;
-- Task 3 --
CREATE OR REPLACE FUNCTION rearrange_salaries() RETURNS VOID AS $$
DECLARE
salaries RECORD;
small NUMERIC[];
larg NUMERIC[];
const CONSTANT INTEGER = 10;
BEGIN
FOR salaries IN
SELECT id, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro
LIMIT const
LOOP
small = array_append(small, (SELECT salary_in_euro FROM bd6_employees WHERE id = salaries.id));
UPDATE bd6_employees SET manager_id = NULL WHERE manager_id = salaries.id;
DELETE FROM bd6_employees
WHERE id = salaries.id;
END LOOP;
FOR salaries IN
SELECT id, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro DESC
LIMIT const
LOOP
larg = array_append(larg, (SELECT id FROM bd6_employees WHERE id = salaries.id));
END LOOP;
FOR i IN 1..const
LOOP
UPDATE bd6_employees
SET salary_in_euro = salary_in_euro + small[i]
WHERE id = larg[i];
END LOOP;
END;
$$ LANGUAGE plpgsql;
SELECT rearrange_salaries();
-- Task 4 --
CREATE OR REPLACE PROCEDURE f() AS $$
DECLARE d_attrs RECORD;
BEGIN
FOR d_attrs IN WITH RECURSIVE tmp AS (
SELECT 1+i*5 AS f1, 2+i*5 AS f2, 3+i*5 AS f3, 4+i*5 AS f4,5+i*5 AS f5
FROM generate_series(0,199) i)
SELECT * FROM tmp
LOOP
case
when ( mod(d_attrs.f1,2)!=0)
then RAISE INFO ' % % % % %', d_attrs.f1, d_attrs.f2, d_attrs.f3, d_attrs.f4, d_attrs.f5;
when ( mod(d_attrs.f1,2)=0)
then RAISE INFO ' % % % % %', d_attrs.f5, d_attrs.f4, d_attrs.f3, d_attrs.f2, d_attrs.f1;
end case;
END LOOP;
END $$ LANGUAGE plpgsql;
call f();
-- Task 7 from lab4 --
DROP VIEW IF EXISTS dep_staff_counts;
CREATE VIEW dep_staff_counts AS
SELECT department, COUNT(*) AS ecount
FROM staff
GROUP BY department;
SELECT * FROM dep_staff_counts;