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ada-dmitry
2023-11-27 13:54:19 +03:00
parent 81ec68f4fa
commit 48c98e32c8
3 changed files with 158 additions and 148 deletions
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Lab/Lab#6/lab6.sql
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WITH RECURSIVE tmp AS (
SELECT id, first_name, last_name, manager_id, 1 AS level
FROM bd6_employees
WHERE manager_id = 1 OR 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;
CREATE OR REPLACE PROCEDURE department_names() AS $$
DECLARE d_attrs RECORD;
BEGIN
FOR d_attrs IN
WITH RECURSIVE tmp AS (
SELECT
id,
first_name,
last_name, manager_id,
1 AS level
FROM bd6_employees
WHERE manager_id = 1 or id = 1
UNION ALL
SELECT
e.id,
e.first_name,
e.last_name,
e.manager_id,
t.level + 1
FROM bd6_employees e
JOIN tmp t ON e.manager_id = t.id )
SELECT * FROM tmp LOOP RAISE INFO ' % % ',
d_attrs.first_name, d_attrs.last_name;
END LOOP;
END $$ LANGUAGE plpgsql;
CALL department_names();
DECLARE
prev_employee NUMBER := 0;
mod_salary NUMBER;
BEGIN
FOR tmp IN (
SELECT first_name, last_name, salary,
ROW_NUMBER() OVER (ORDER BY salary) AS salary_rank
FROM employees
)
LOOP
IF emp.salary_rank = 1 THEN
modified_salary := FLOOR(emp.salary / 100) * 100; -- округляем до сотен в меньшую сторону
ELSE
modified_salary := FLOOR((emp.salary + previous_remainder) / 100) * 100;
previous_remainder := (emp.salary + previous_remainder) - modified_salary;
END IF;
DBMS_OUTPUT.PUT_LINE('Фамилия: ' || emp.last_name || ', Имя: ' || emp.first_name || ', Модифицированная заработная плата: ' || modified_salary);
END LOOP;
END;
CREATE OR REPLACE PROCEDURE f() AS $$
DECLARE d_attrs RECORD;
BEGIN
FOR d_attrs IN
WITH RECURSIVE tmp AS (
SELECT i AS f1, i2 AS f2, i3 AS f3,i4 AS f4,i5 AS f5
FROM generate_series(1,200) i)
SELECT * FROM tmp LOOP RAISE INFO ' % % % % %',
d_attrs.f1, d_attrs.f2, d_attrs.f3, d_attrs.f4, d_attrs.f5; END LOOP;
END $$ LANGUAGE plpgsql;
CALL f();
WITH RECURSIVE tmp AS (
SELECT id, first_name, last_name, manager_id, 1 AS level
FROM bd6_employees
WHERE manager_id = 1 OR 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;
CREATE OR REPLACE PROCEDURE department_names() AS $$
DECLARE d_attrs RECORD;
BEGIN
FOR d_attrs IN
WITH RECURSIVE tmp AS (
SELECT
id,
first_name,
last_name, manager_id,
1 AS level
FROM bd6_employees
WHERE manager_id = 1 or id = 1
UNION ALL
SELECT
e.id,
e.first_name,
e.last_name,
e.manager_id,
t.level + 1
FROM bd6_employees e
JOIN tmp t ON e.manager_id = t.id )
SELECT * FROM tmp LOOP RAISE INFO ' % % ',
d_attrs.first_name, d_attrs.last_name;
END LOOP;
END $$ LANGUAGE plpgsql;
CALL department_names();
CREATE OR REPLACE PROCEDURE modified_salary() AS $$
DECLARE
prev_employee numeric := 0;
mod_salary numeric := (
SELECT FLOOR(min(salary_in_euro) / 100) * 100
FROM bd6_employees
);
d_attrs RECORD;
BEGIN
FOR d_attrs IN WITH tmp AS (
SELECT first_name, last_name, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro ASC
)
SELECT * FROM tmp
RAISE INFO ' % % % ',
d_attrs.last_name,
d_attrs.first_name,
mod_salary;
LOOP
mod_salary := FLOOR((d_attrs.salary_in_euro + prev_employee) / 100) * 100;
prev_employee := (d_attrs.salary_in_euro + prev_employee) - mod_salary;
RAISE INFO ' % % % ',
d_attrs.last_name,
d_attrs.first_name,
mod_salary;
END LOOP;
END $$ LANGUAGE plpgsql;
CALL modified_salary();
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. */
CREATE OR REPLACE PROCEDURE department_names() AS
DECLARE d_attrs RECORD;
BEGIN
FOR d_attrs IN WITH RECURSIVE tmp AS (
SELECT id, first_name, last_name, manager_id, 1 AS level
FROM bd6_employees
WHERE manager_id = 1 or 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
LOOP
RAISE INFO ' % % ', d_attrs.first_name, d_attrs.last_name;
END LOOP;
END LANGUAGE plpgsql;
CALL department_names();
-- 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. */
CREATE OR REPLACE PROCEDURE print_employees() RETURNS VOID AS $$
DECLARE
current_salary INTEGER;
m_salary INTEGER;
mod_salary INTEGER := 0;
employee_rec RECORD;
BEGIN
FOR employee_rec IN (
SELECT last_name, first_name, salary_in_euro
FROM bd6_employees
ORDER BY salary_in_euro)
LOOP
current_salary := employee_rec.salary_in_euro;
IF mod_salary=0 THEN
m_salary := (current_salary / 100) * 100;
ELSE
m_salary := ((current_salary + mod_salary) / 100) * 100;
END IF;
RAISE NOTICE 'Employee: % %, Modified Salary: %', employee_rec.last_name, employee_rec.first_name, m_salary;
mod_salary := current_salary - m_salary;
END LOOP;
END $$ LANGUAGE plpgsql;
SELECT print_employees();
-- 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 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();
-- 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 $$
DECLARE d_attrs RECORD;
BEGIN
FOR d_attrs IN WITH RECURSIVE tmp AS (
SELECT 1+i5 AS f1, 2+i5 AS f2, 3+i5 AS f3, 4+i5 AS f4,5+i*5 AS f5
FROM generate_series(0,199) i)
SELECT * FROM tmp
LOOP
case
when ( mod(d_attrs.f1,2)!=0)
then RAISE INFO ' % % % % %', d_attrs.f1, d_attrs.f2, d_attrs.f3, d_attrs.f4, d_attrs.f5;
when ( mod(d_attrs.f1,2)=0)
then RAISE INFO ' % % % % %', d_attrs.f5, d_attrs.f4, d_attrs.f3, d_attrs.f2, d_attrs.f1;
end case;
END LOOP;
END $$ LANGUAGE plpgsql;
call f();
Lab/Lab#6/test3.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();