Dmitry Medvedev
230 lines
6.7 KiB
SQL
230 lines
6.7 KiB
SQL
# Задачи на отработку:
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### **Задача 1.**
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Есть таблица `users`:
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| user_id | name | age | country | is_active |
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| --- | --- | --- | --- | --- |
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| 1 | Ivan | 30 | RU | 1 |
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| 2 | Anna | 25 | RU | 1 |
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| 3 | Petr | 45 | RU | 0 |
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| 4 | Olga | 35 | RU | 1 |
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| 5 | John | 55 | US | 1 |
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| 6 | Alice | 42 | US | 0 |
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| 7 | Tom | 38 | US | 1 |
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| 8 | Lucas | 28 | BR | 1 |
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| 9 | Maria | 60 | BR | 1 |
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| 10 | João | 27 | BR | 0 |
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| 11 | Kenji | 33 | JP | 1 |
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| 12 | Yuki | 29 | JP | 0 |
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| 13 | Chen | 40 | CN | 1 |
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| 14 | Li | 48 | CN | 1 |
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| 15 | Fatima | 31 | AE | 1 |
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| 16 | Ahmed | 36 | AE | 1 |
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| 17 | George | 24 | US | 1 |
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| 18 | Nina | 34 | RU | 1 |
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| 19 | Max | 50 | DE | 1 |
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| 20 | Sophie | 22 | DE | 0 |
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CREATE TABLE users (
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user_id UInt64,
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name String,
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age UInt32,
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country String,
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is_active UInt32
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)
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ENGINE = MergeTree()
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ORDER BY (age, country);
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1. Получи 10 самых старших **активных** пользователей.
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SELECT
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user_id,
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name,
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age
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FROM users
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WHERE is_active = 1
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ORDER BY age DESC
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LIMIT 10;
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2. Выведи пользователей из `country = 'RU'`, отсортированных по имени.
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SELECT
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user_id,
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name
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FROM users
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WHERE country = 'RU'
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ORDER BY name;
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3. Что произойдёт, если написать `LIMIT 10` **без `ORDER BY`**?
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В таком случае CH покажет 10 произвольных неотсортированных строк
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4. Посчитай количество пользователей по странам.
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SELECT
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country,
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COUNT(*)
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FROM users
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GROUP BY country;
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5. Для каждой страны выведи минимальный и максимальный возраст.
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SELECT
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country,
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MIN(age),
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MAX(age)
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FROM users
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GROUP BY country;
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6. Отсортируй страны по убыванию количества пользователей.
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SELECT
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country,
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COUNT(*)
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FROM users
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GROUP BY country
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ORDER BY COUNT(*) DESC;
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📌 Дополнительно:
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- Используй `count()`, `min()`, `max()`, `sum()`
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- Попробуй `uniqExact(country)` — что он делает?
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uniqExact выводит количество уникальных значений, это тоже самое что и DISTINCT
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### **Задача 2.**
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Есть таблица `events`:
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| user_id | event_time | event_type |
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| --- | --- | --- |
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| 1 | 2025-07-17 09:00:00 | click |
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| 1 | 2025-07-17 09:05:00 | view |
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| 1 | 2025-07-17 09:10:00 | purchase |
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| 2 | 2025-07-17 10:00:00 | click |
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| 2 | 2025-07-17 10:05:00 | click |
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| 2 | 2025-07-17 11:00:00 | view |
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| 3 | 2025-07-16 12:00:00 | view |
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| 3 | 2025-07-16 12:30:00 | click |
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| 3 | 2025-07-17 08:00:00 | purchase |
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| 4 | 2025-07-17 13:00:00 | click |
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| 4 | 2025-07-17 14:00:00 | view |
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| 4 | 2025-07-17 15:00:00 | view |
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| 5 | 2025-07-16 20:00:00 | view |
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| 5 | 2025-07-17 01:00:00 | click |
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| 5 | 2025-07-17 06:00:00 | purchase |
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| 6 | 2025-07-17 07:30:00 | view |
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| 6 | 2025-07-17 08:45:00 | click |
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| 7 | 2025-07-16 23:59:59 | click |
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| 7 | 2025-07-17 00:01:00 | purchase |
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| 7 | 2025-07-17 10:10:00 | view |
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CREATE TABLE events (
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user_id UInt64,
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event_type String,
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event_time DateTime64(3)
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)
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ENGINE = MergeTree()
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ORDER BY (event_time, user_id);
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INSERT INTO dmedvedev.events (user_id, event_time, event_type) VALUES
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(1, '2025-07-17 09:00:00', 'click'),
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(1, '2025-07-17 09:05:00', 'view'),
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(1, '2025-07-17 09:10:00', 'purchase'),
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(2, '2025-07-17 10:00:00', 'click'),
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(2, '2025-07-17 10:05:00', 'click'),
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(2, '2025-07-17 11:00:00', 'view'),
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(3, '2025-07-16 12:00:00', 'view'),
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(3, '2025-07-16 12:30:00', 'click'),
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(3, '2025-07-17 08:00:00', 'purchase'),
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(4, '2025-07-17 13:00:00', 'click'),
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(4, '2025-07-17 14:00:00', 'view'),
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(4, '2025-07-17 15:00:00', 'view'),
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(5, '2025-07-16 20:00:00', 'view'),
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(5, '2025-07-17 01:00:00', 'click'),
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(5, '2025-07-17 06:00:00', 'purchase'),
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(6, '2025-07-17 07:30:00', 'view'),
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(6, '2025-07-17 08:45:00', 'click'),
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(7, '2025-07-16 23:59:59', 'click'),
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(7, '2025-07-17 00:01:00', 'purchase'),
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(7, '2025-07-17 10:10:00', 'view');
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1. Добавь колонку с кумулятивным количеством событий по `user_id` (используй `count() OVER ...`).
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SELECT
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user_id,
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COUNT() OVER (PARTITION BY user_id ORDER BY user_id)
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FROM events;
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2. Сделай аналог ранжирования событий внутри пользователя: `row_number() OVER (PARTITION BY user_id ORDER BY event_time)`
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SELECT
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user_id,
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COUNT(*) OVER (PARTITION BY user_id ORDER BY event_time)
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FROM events;
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3. Что будет, если ты явно задашь `ROWS BETWEEN`? Поддерживается ли?
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Посчитается, поддерживается.
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SELECT
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user_id,
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COUNT(*) OVER (
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PARTITION BY user_id ORDER BY event_time
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ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
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)
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FROM events;
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4. Сделай `WITH recent_events AS (...)` — все события за последние 24 часа.
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SELECT
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user_id,
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event_time
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FROM events
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WHERE event_time <= NOW() - INTERVAL 24 HOUR;
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5. Используй CTE, чтобы посчитать топ-5 пользователей по числу этих событий.
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WITH
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recent_events AS (
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SELECT
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user_id,
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event_type,
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event_time
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FROM events
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WHERE event_time >= NOW() - INTERVAL 24 HOUR
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)
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SELECT
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user_id,
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COUNT(*)
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FROM recent_events
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GROUP BY user_id
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ORDER BY COUNT(*) DESC
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LIMIT 5;
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6. Добавь второй `WITH` — для маппинга `event_type → category` и используй его в основном запросе.
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WITH
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recent_events AS (
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SELECT
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user_id,
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event_type,
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event_time
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FROM events
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WHERE event_time >= NOW() - INTERVAL 24 HOUR
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),
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top_users AS(
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SELECT
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user_id,
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COUNT(*)
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FROM recent_events
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GROUP BY user_id
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ORDER BY COUNT(*) DESC
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LIMIT 5
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)
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SELECT
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user_id,
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CASE
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WHEN event_type IN ('click', 'view') THEN 'interaction'
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WHEN event_type IN ('purchase') THEN 'conversion'
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ELSE 'unknown'
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END AS category
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FROM recent_events re INNER JOIN top_users tu ON re.user_id = tu.user_id;
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📌 Вопрос: поддерживает ли ClickHouse **многоуровневые** или **рекурсивные** CTE?
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МНОГОУРОВНЕВОСТЬ ЕСТЬ, А РЕКУРСИВНОСТИ НЕТ. |