Что такое запрос?sql-8

SELECT employee_id, first_name, last_name
FROM employees
WHERE department = 'IT'
ORDER BY hire_date DESC;

-- Вставка
INSERT INTO products (name, price)
VALUES ('Ноутбук', 999.99);

-- Обновление
UPDATE customers
SET loyalty_points = loyalty_points + 100
WHERE last_purchase_date > '2023-01-01';

-- Удаление
DELETE FROM temp_logs
WHERE created_at < CURRENT_DATE - INTERVAL '30 days';

CREATE TABLE orders (
    order_id SERIAL PRIMARY KEY,
    customer_id INT REFERENCES customers(customer_id),
    order_date DATE NOT NULL DEFAULT CURRENT_DATE
);

ALTER TABLE employees
ADD COLUMN email VARCHAR(255) UNIQUE;

DROP TABLE IF EXISTS backup_data;

SELECT
    d.department_name,
    COUNT(e.employee_id) AS employee_count,
    AVG(e.salary) AS avg_salary
FROM
    employees e
JOIN
    departments d ON e.department_id = d.department_id
WHERE
    e.hire_date > '2020-01-01'
GROUP BY
    d.department_name
HAVING
    COUNT(e.employee_id) > 5
ORDER BY
    avg_salary DESC
LIMIT 10;

SELECT * FROM products WHERE price < 100;

SELECT o.order_id, c.name, p.product_name
FROM orders o
JOIN customers c ON o.customer_id = c.customer_id
JOIN order_items oi ON o.order_id = oi.order_id
JOIN products p ON oi.product_id = p.product_id;

SELECT name
FROM employees
WHERE department_id IN (
    SELECT department_id
    FROM departments
    WHERE location = 'Москва'
);

WITH RECURSIVE employee_hierarchy AS (
    SELECT employee_id, name, manager_id
    FROM employees
    WHERE employee_id = 100  -- Начало с CEO

    UNION ALL

    SELECT e.employee_id, e.name, e.manager_id
    FROM employees e
    JOIN employee_hierarchy eh ON e.manager_id = eh.employee_id
)
SELECT * FROM employee_hierarchy;

CREATE INDEX idx_employee_department ON employees(department_id);

EXPLAIN ANALYZE
SELECT * FROM large_table WHERE category = 'A';

SELECT /*+ HASH_JOIN(orders, customers) */
    o.order_id, c.name
FROM orders o
JOIN customers c ON o.customer_id = c.customer_id;