Примеры
На этой странице собраны практические примеры использования SQL Prettify для типовых SQL-запросов. Блоки SQL сохранены без изменений, чтобы было проще сравнивать исходный и отформатированный вариант.
Базовые запросы SELECT
Простой SELECT
Вход:
select id, name, email from users where active = 1;
Форматированный результат:
SELECT
id,
name,
email
FROM users
WHERE active = 1;
SELECT с ORDER BY
Вход:
select u.id, u.name, u.email from users u where u.active = 1 order by u.name asc, u.created_at desc;
Форматированный результат:
SELECT
u.id,
u.name,
u.email
FROM users u
WHERE u.active = 1
ORDER BY
u.name ASC,
u.created_at DESC;
Запросы с JOIN
INNER JOIN
Вход:
select u.id, u.name, u.email, p.title, p.content from users u inner join posts p on u.id = p.user_id where u.active = 1 and p.published = 1;
Форматированный результат:
SELECT
u.id,
u.name,
u.email,
p.title,
p.content
FROM users u
INNER JOIN posts p ON u.id = p.user_id
WHERE u.active = 1
AND p.published = 1;
Агрегационные запросы
GROUP BY с агрегацией
Вход:
select u.id, u.name, count(p.id) as post_count, sum(p.views) as total_views from users u left join posts p on u.id = p.user_id where u.active = 1 group by u.id, u.name having count(p.id) > 0 order by total_views desc;
Форматированный результат:
SELECT
u.id,
u.name,
COUNT(p.id) AS post_count,
SUM(p.views) AS total_views
FROM users u
LEFT JOIN posts p ON u.id = p.user_id
WHERE u.active = 1
GROUP BY
u.id,
u.name
HAVING COUNT(p.id) > 0
ORDER BY total_views DESC;
Подзапросы
Подзапрос в WHERE
Вход:
select id, name, email from users where id in (select user_id from posts where published = 1) and active = 1;
Форматированный результат:
SELECT
id,
name,
email
FROM users
WHERE id IN (
SELECT user_id
FROM posts
WHERE published = 1
)
AND active = 1;
Коррелированный подзапрос
Вход:
select u.id, u.name, u.email, (select count(*) from posts p where p.user_id = u.id and p.published = 1) as post_count from users u where u.active = 1;
Форматированный результат:
SELECT
u.id,
u.name,
u.email,
(
SELECT COUNT(*)
FROM posts p
WHERE p.user_id = u.id
AND p.published = 1
) AS post_count
FROM users u
WHERE u.active = 1;
Изменение данных
Оператор INSERT
Вход:
insert into users (name, email, active, created_at) values ('John Doe', 'john@example.com', 1, now()), ('Jane Smith', 'jane@example.com', 1, now());
Форматированный результат:
INSERT INTO users (
name,
email,
active,
created_at
)
VALUES (
'John Doe',
'john@example.com',
1,
NOW()
), (
'Jane Smith',
'jane@example.com',
1,
NOW()
);
Оператор UPDATE
Вход:
update users set name = 'John Updated', email = 'john.updated@example.com', updated_at = now() where id = 1 and active = 1;
Форматированный результат:
UPDATE users
SET
name = 'John Updated',
email = 'john.updated@example.com',
updated_at = NOW()
WHERE id = 1
AND active = 1;
Оператор DELETE
Вход:
delete from posts where user_id in (select id from users where active = 0) and published = 0;
Форматированный результат:
DELETE FROM posts
WHERE user_id IN (
SELECT id
FROM users
WHERE active = 0
)
AND published = 0;
Определение данных
CREATE TABLE
Вход:
create table users (id int primary key auto_increment, name varchar(255) not null, email varchar(255) unique not null, active boolean default true, created_at timestamp default current_timestamp, updated_at timestamp default current_timestamp on update current_timestamp);
Форматированный результат:
CREATE TABLE users (
id INT PRIMARY KEY AUTO_INCREMENT,
name VARCHAR(255) NOT NULL,
email VARCHAR(255) UNIQUE NOT NULL,
active BOOLEAN DEFAULT TRUE,
created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
updated_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP
);
Оконные функции
ROW_NUMBER()
Вход:
select id, name, email, row_number() over (partition by department_id order by created_at desc) as row_num from users where active = 1;
Форматированный результат:
SELECT
id,
name,
email,
ROW_NUMBER() OVER (
PARTITION BY department_id
ORDER BY created_at DESC
) AS row_num
FROM users
WHERE active = 1;
Common Table Expressions (CTE)
Конструкция WITH
Вход:
with active_users as (select id, name, email from users where active = 1), user_posts as (select user_id, count(*) as post_count from posts where published = 1 group by user_id) select au.id, au.name, au.email, coalesce(up.post_count, 0) as post_count from active_users au left join user_posts up on au.id = up.user_id order by post_count desc;
Форматированный результат:
WITH active_users AS (
SELECT
id,
name,
email
FROM users
WHERE active = 1
),
user_posts AS (
SELECT
user_id,
COUNT(*) AS post_count
FROM posts
WHERE published = 1
GROUP BY user_id
)
SELECT
au.id,
au.name,
au.email,
COALESCE(up.post_count, 0) AS post_count
FROM active_users au
LEFT JOIN user_posts up ON au.id = up.user_id
ORDER BY post_count DESC;
Параметры форматирования
Разные размеры отступов
2 пробела:
SELECT
id,
name,
email
FROM users
WHERE active = 1;
4 пробела:
SELECT
id,
name,
email
FROM users
WHERE active = 1;
Разный регистр ключевых слов
ВЕРХНИЙ РЕГИСТР:
SELECT
id,
name,
email
FROM users
WHERE active = 1;
нижний регистр:
select
id,
name,
email
from users
where active = 1;
Эти примеры показывают, как SQL Prettify улучшает читаемость запросов и помогает поддерживать единый стиль в команде.