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3 sem/Lab/Lab#6/bd6_inserts.sql
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DROP TABLE IF EXISTS bd6_departments;
DROP TABLE IF EXISTS bd6_employees;
CREATE TABLE bd6_departments(
id integer PRIMARY KEY,
name varchar(64),
postal_code varchar(6),
street varchar(64),
building varchar(16),
city varchar(32)
);
INSERT INTO bd6_departments
VALUES(10, 'Administration', '109658', 'Leningradskoe shosse', '1', 'Moscow'),
(20, 'Marketing', '107701', 'Lenina', '22a', 'Volgograd'),
(30, 'Purchasing', '109901', 'Mikluho-Maklaya', '8', 'Bryansk'),
(40, 'Human Resources', '10967', '5-ya parkovaya', '16', 'Moscow'),
(50, 'Shipping', '109659', '38 Bakinskih komissarov', '77', 'Moscow'),
(60, 'IT', '109902', 'Pervomajskaya', '33', 'Kirov');
CREATE TABLE bd6_employees(
id integer PRIMARY KEY,
last_name varchar(64) NOT NULL,
first_name varchar(64) NOT NULL,
phone_number varchar(18),
email varchar(32),
department_id integer NOT NULL,
manager_id integer,
salary_in_euro numeric(8, 2) DEFAULT 0 NOT NULL,
UNIQUE(last_name, first_name, department_id)
);
INSERT INTO bd6_employees VALUES(1, 'Radygin', 'Victor', '8-(495)-567-7788', 'vr@e.mail.mephi.ru', 10, NULL, 8000),
(2, 'Kuprijanov', 'Dmitrij', '8-(495)-567-7794', 'kd@e.mail.mephi.ru', 60, 1, 6534.33),
(3, 'Ivanov-Skladovskij', 'Ivan', '8-(495)-567-7799', 'ii1@mail.mephi.ru', 20, 1, 4404.14),
(4, 'Petrov', 'Petr', '8-(495)-567-7794', 'petrovpetr@m.gmail.ru', 60, 2, 3456.43),
(5, 'kozlov', 'Konstantin', '8-(495)-567-7794', 'kkozlov@mephi.ru', 60, 2, 2300),
(6, 'Abramov', 'Abram', '8-(495)-567-7794', 'abramova@k75.mephi.ru', 60, 2, 2200.11),
(7, 'IvanovпїЅ-SkladovskпїЅya-Petrova', 'Ivanka', '8-(495)-567-7794', 'ii2@mail.mephi.ru', 60, 4, 3756.33),
(8, 'Petrov', 'Ivan', '8-(495)-567-7799', 'petrovivan@m.gmail.ru', 20, 3, 4850),
(9, 'kozlov', 'Ivan', '8-(495)-567-7799', 'ikozlov@mephi.ru', 20, 3, 3460),
(10, 'Abramov', 'Moisej', '8-(495)-567-7794', 'abramovm@k75.mephi.ru', 60, 4, 2345),
(11, 'Petrov', 'Alex', '8-(495)-567-7799', 'petroval@m.gmail.ru', 20, 3, 2465),
(12, 'kozlov', 'Maxim', '8-(495)-567-7799', 'mkozlov@mephi.ru', 20, 3, 2788),
(13, 'Abramov', 'Isyaslav', '8-(495)-567-7738', 'abramovi@k75.mephi.ru', 30, 1, 6500);
3 sem/Lab/Lab#6/insert.sql
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DROP TABLE IF EXISTS bd6_employees;
CREATE TABLE bd6_departments(
id integer PRIMARY KEY,
name varchar(64),
postal_code varchar(6),
street varchar(64),
building varchar(16),
city varchar(32)
);
CREATE TABLE
bd6_employees (
id integer PRIMARY KEY,
last_name varchar(64) NOT NULL,
first_name varchar(64) NOT NULL,
phone_number varchar(18),
email varchar(32),
department_id integer REFERENCES bd6_departments (id) NOT NULL,
manager_id integer REFERENCES bd6_employees (id),
salary_in_euro numeric(8, 2) DEFAULT 0 NOT NULL,
UNIQUE (last_name, first_name, department_id)
);
INSERT INTO bd6_departments
VALUES(10, 'Administration', '109658', 'Leningradskoe shosse', '1', 'Moscow'),
(20, 'Marketing', '107701', 'Lenina', '22a', 'Volgograd'),
(30, 'Purchasing', '109901', 'Mikluho-Maklaya', '8', 'Bryansk'),
(40, 'Human Resources', '10967', '5-ya parkovaya', '16', 'Moscow'),
(50, 'Shipping', '109659', '38 Bakinskih komissarov', '77', 'Moscow'),
(60, 'IT', '109902', 'Pervomajskaya', '33', 'Kirov');
INSERT INTO
bd6_employees
VALUES
(
1,
'Radygin',
'Victor',
'8-(495)-567-7788',
'vr@e.mail.mephi.ru',
10,
NULL,
8000
),
(
2,
'Kuprijanov',
'Dmitrij',
'8-(495)-567-7794',
'kd@e.mail.mephi.ru',
60,
1,
6534.33
),
(
3,
'Ivanov-Skladovskij',
'Ivan',
'8-(495)-567-7799',
'ii1@mail.mephi.ru',
20,
1,
4404.14
),
(
4,
'Petrov',
'Petr',
'8-(495)-567-7794',
'petrovpetr@m.gmail.ru',
60,
2,
3456.43
),
(
5,
'kozlov',
'Konstantin',
'8-(495)-567-7794',
'kkozlov@mephi.ru',
60,
2,
2300
),
(
6,
'Abramov',
'Abram',
'8-(495)-567-7794',
'abramova@k75.mephi.ru',
60,
2,
2200.11
),
(
7,
'IvanovпїЅ-SkladovskпїЅya-Petrova',
'Ivanka',
'8-(495)-567-7794',
'ii2@mail.mephi.ru',
60,
4,
3756.33
),
(
8,
'Petrov',
'Ivan',
'8-(495)-567-7799',
'petrovivan@m.gmail.ru',
20,
3,
4850
),
(
9,
'kozlov',
'Ivan',
'8-(495)-567-7799',
'ikozlov@mephi.ru',
20,
3,
3460
),
(
10,
'Abramov',
'Moisej',
'8-(495)-567-7794',
'abramovm@k75.mephi.ru',
60,
4,
2345
),
(
11,
'Petrov',
'Alex',
'8-(495)-567-7799',
'petroval@m.gmail.ru',
20,
3,
2465
),
(
12,
'kozlov',
'Maxim',
'8-(495)-567-7799',
'mkozlov@mephi.ru',
20,
3,
2788
),
(
13,
'Abramov',
'Isyaslav',
'8-(495)-567-7738',
'abramovi@k75.mephi.ru',
30,
1,
6500
);
3 sem/Lab/Lab#6/lab6.sql
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CREATE OR REPLACE PROCEDURE update_salary() AS $$
DECLARE
emp_record bd6_employees%ROWTYPE;
counter integer := 0;
min_salary bd6_employees.salary_in_euro%TYPE;
max_salary bd6_employees.salary_in_euro%TYPE;
BEGIN
-- SELECT MIN(salary_in_euro), MAX(salary_in_euro) INTO min_salary, max_salary FROM bd6_employees;
FOR emp_record IN
SELECT * FROM (SELECT * FROM bd6_employees ORDER BY salary_in_euro LIMIT 3) AS min_salaries
LOOP
DELETE FROM bd6_employees WHERE id = emp_record.id;
counter := counter + 1;
END LOOP;
FOR emp_record IN
SELECT * FROM (SELECT * FROM bd6_employees ORDER BY salary_in_euro DESC LIMIT 3) AS max_salaries
LOOP
UPDATE bd6_employees SET salary_in_euro = salary_in_euro + min_salary WHERE id = emp_record.id;
counter := counter + 1;
END LOOP;
RAISE NOTICE 'Successfully updated % bd6_employees.', counter;
END $$ LANGUAGE plpgsql;
CALL update_salary();
CREATE OR REPLACE PROCEDURE update_and_delete_salaries() AS $$
BEGIN
-- Увеличение заработной платы
WITH lowest_salaries AS (
SELECT id, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro
LIMIT 10
), highest_salary AS (
SELECT max(salary_in_euro) AS max_salary
FROM bd6_employees
)
UPDATE bd6_employees e
SET salary_in_euro = e.salary_in_euro + ls.salary_in_euro - (SELECT max_salary FROM highest_salary)
FROM lowest_salaries ls
WHERE e.id = ls.id;
-- Удаление сотрудников
DELETE FROM bd6_employees WHERE id IN (SELECT id FROM lowest_salaries);
RAISE NOTICE 'Salaries updated and lowest paid bd6_employees deleted.';
END;
$$ LANGUAGE plpgsql;
3 sem/Lab/Lab#6/lab6_done.sql
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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;
3 sem/Lab/Lab#6/test3.sql
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CREATE OR REPLACE FUNCTION rearrange_salaries() RETURNS VOID AS $$
DECLARE
salaries RECORD;
small NUMERIC[];
larg NUMERIC[];
const CONSTANT INTEGER = 10;
BEGIN
FOR salaries IN
SELECT id, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro
LIMIT const
LOOP
small = array_append(small, (SELECT salary_in_euro FROM bd6_employees WHERE id = salaries.id));
DELETE FROM bd6_employees
WHERE id = salaries.id;
END LOOP;
FOR salaries IN
SELECT id, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro DESC
LIMIT const
LOOP
larg = array_append(larg, (SELECT id FROM bd6_employees WHERE id = salaries.id));
END LOOP;
FOR i IN 1..const
LOOP
UPDATE bd6_employees
SET salary_in_euro = salary_in_euro + small[i]
WHERE id = larg[i];
END LOOP;
END;
$$ LANGUAGE plpgsql;
SELECT rearrange_salaries();
CREATE OR REPLACE FUNCTION update_salaries() RETURNS VOID AS $$
DECLARE
max_salary_employee_id INTEGER;
min_salaries numeric[];
max_salaries numeric[];
i INTEGER;
const CONSTANT INTEGER = 10;
BEGIN
-- Получаем заработные платы последних 10 человек с самой маленькой заработной платой
SELECT array_agg(id) FROM bd6_employees
ORDER BY salary_in_euro ASC
LIMIT const INTO min_salaries;
-- Получаем заработные платы 10 первых человек с самой большой заработной платой
SELECT array_agg(id) FROM bd6_employees
ORDER BY salary_in_euro DESC
LIMIT const INTO max_salaries;
-- Добавляем заработные платы
FOR i IN 1..const LOOP
UPDATE bd6_employees
SET salary_in_euro = salary_in_euro + min_salaries[i]
WHERE id = max_salaries[i];
END LOOP;
-- Удаляем 10 человек с наименьшей заработной платой
DELETE FROM bd6_employees
WHERE id IN (SELECT unnest(min_salaries));
RAISE NOTICE 'Заработные платы обновлены и % человек удалены', const;
END;
$$ LANGUAGE plpgsql;