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SELECT JSON_ARRAYAGG(category_result) AS films_by_category FROM ( SELECT JSON_OBJECT(c.name, COUNT(f.film_id)) AS category_result FROM category c JOIN film_category fc ON c.category_id = fc.category_id JOIN film f ON fc.film_id = f.film_id GROUP BY c.category_id ) AS sub; -- Используйте агрегатную функцию JSON_ARRAYAGG и функцию JSON_OBJECT.

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