## Контекст (таблицы и пример данных) ### Таблицы - `users(id, full_name, email, city, age, registe
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## Контекст (таблицы и пример данных)
### Таблицы
- `users(id, full_name, email, city, age, registered_at)`
- `orders(id, user_id, order_date, amount, status)`
- `payments(id, order_id, payment_date, payment_type, paid_amount)`
### Пример данных
**users**
| id | full_name | email | city | age | registered_at |
|---:|----------------|---------------------|----------|----:|---------------|
| 1 | Ivan Petrov | ivan@mail.com | Helsinki | 29 | 2025-01-10 |
| 2 | Anna Ivanova | anna@gmail.com | Espoo | 34 | 2025-02-01 |
| 3 | Ivan Petrov 2 | ivan@mail.com | Vantaa | 41 | 2025-02-05 |
**orders**
| id | user_id | order_date | amount | status |
|----:|--------:|-------------|--------:|--------|
| 10 | 1 | 2025-03-01 | 120.00 | paid |
| 11 | 2 | 2025-03-02 | 2500.00 | new |
| 12 | 1 | 2025-03-05 | 80.00 | paid |
**payments**
| id | order_id | payment_date | payment_type | paid_amount |
|----:|---------:|-------------|--------------|------------:|
| 100 | 10 | 2025-03-01 | card | 120.00 |
| 101 | 11 | 2025-03-03 | transfer | 1000.00 |
| 102 | 11 | 2025-03-04 | transfer | 1500.00 |
---
## Задача
Найти пользователей с одинаковым `email`.
Необходимо вывести `email` и количество повторений.
Ответы
```sql
SELECT email, COUNT(*) AS cnt
FROM users
GROUP BY email
HAVING COUNT(*) > 1;
```