Как обновить таблицу в базе данных с помощью SQL?sql-14

UPDATE table_name
SET column1 = value1,
    column2 = value2, ...
[WHERE condition];

UPDATE employees
SET salary = 50000,
    department = 'IT'
WHERE employee_id = 101;

UPDATE products
SET price = 29.99
WHERE product_id = 1001;

UPDATE customers
SET first_name = 'John',
    last_name = 'Smith',
    email = 'john.smith@example.com'
WHERE customer_id = 42;

UPDATE orders
SET total_amount = quantity * unit_price * (1 - discount)
WHERE order_id = 12345;

UPDATE employees
SET salary = salary * 1.1
WHERE department_id IN (
    SELECT department_id
    FROM departments
    WHERE location = 'New York'
);

-- ОПАСНО: обновит все строки в таблице!
UPDATE products SET discontinued = 1;

-- Безопасный вариант
UPDATE products SET discontinued = 1
WHERE product_id = 1001;

-- Сначала посмотреть, что будет обновлено
SELECT * FROM employees
WHERE salary < 30000;

-- Затем выполнить обновление
UPDATE employees
SET salary = salary * 1.15
WHERE salary < 30000;

BEGIN TRANSACTION;

UPDATE accounts
SET balance = balance - 100
WHERE account_id = 1;

UPDATE accounts
SET balance = balance + 100
WHERE account_id = 2;

-- Проверить изменения перед коммитом
-- ROLLBACK; -- если что-то не так
COMMIT;

UPDATE p
SET p.price = p.price * 1.1
FROM products p
JOIN product_categories pc ON p.category_id = pc.category_id
WHERE pc.category_name = 'Electronics';

WITH top_products AS (
    SELECT product_id
    FROM sales
    GROUP BY product_id
    ORDER BY SUM(quantity) DESC
    LIMIT 10
)
UPDATE products
SET featured = 1
WHERE product_id IN (SELECT product_id FROM top_products);

-- Обновить 100 записей за раз (SQL Server)
UPDATE TOP (100) customers
SET last_contact_date = GETDATE()
WHERE last_contact_date IS NULL;

UPDATE employees
SET vacation_days = vacation_days + 5
OUTPUT
    inserted.employee_id,
    deleted.vacation_days AS old_value,
    inserted.vacation_days AS new_value
WHERE hire_date < '2020-01-01';

UPDATE employees e, departments d
SET e.salary = e.salary * 1.1
WHERE e.department_id = d.department_id
AND d.location = 'London';

UPDATE employees
SET salary = salary * 1.1
FROM departments
WHERE employees.department_id = departments.department_id
AND departments.location = 'London';

UPDATE
    (SELECT e.salary, d.location
     FROM employees e
     JOIN departments d ON e.department_id = d.department_id
     WHERE d.location = 'London')
SET salary = salary * 1.1;

BEGIN TRANSACTION;

-- Сначала проверить, какие строки будут затронуты
SELECT * FROM table WHERE condition;

-- Затем выполнить обновление
UPDATE table SET column = value WHERE condition;

-- Проверить результат
SELECT * FROM table WHERE condition;

-- COMMIT или ROLLBACK в зависимости от результата
COMMIT TRANSACTION;