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