homework "6) Агрегации и группировки" done

2026-02-14 13:40:11 +01:00
parent 53d802bb89
commit 65fc8315c5
2 changed files with 122 additions and 0 deletions
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 <?xml version="1.0" encoding="UTF-8"?>
 <projectDescription>
 	<name>postgresql</name>
 	<comment></comment>
 	<projects>
 	</projects>
 	<buildSpec>
 	</buildSpec>
 	<natures>
 	</natures>
 </projectDescription>
@@ -0,0 +1,111 @@
 /* Задание 1 */
 SELECT 
 	category,
 	count(*) AS products_count 
 FROM 
 	products p
 GROUP BY category 
 ORDER BY products_count DESC;
 /* Задание 2
 * Тут всё решение в задании 
 *  */
 SELECT 
 	order_id,
 	sum(quantity) AS total_quantity,
 	sum(quantity * price_at_purchase) AS order_revenue
 FROM 
 	order_items oi 
 GROUP BY order_id 
 ORDER BY order_revenue DESC;
 /* Задание 3 */
 SELECT 
 	p.category,
 	COALESCE(sum(oi.quantity * oi.price_at_purchase), 0) AS category_revenue
 FROM
 	products p
 LEFT JOIN
 	order_items oi ON p.id = oi.product_id 
 GROUP BY p.category 
 /* Задание 4 */
 SELECT 
 	o.id AS order_id,
 	sum(oi.quantity * oi.price_at_purchase) AS order_revenue 
 FROM
 	orders o 
 LEFT JOIN
 	order_items oi ON o.id = oi.order_id  
 WHERE o.status = 'paid'
 GROUP BY 1
 ORDER BY 2 DESC 
 /* Задание 5 */
 SELECT 
 	p.category,
 	COALESCE(sum(oi.quantity * oi.price_at_purchase), 0) AS category_revenue
 FROM
 	products p
 LEFT JOIN
 	order_items oi ON p.id = oi.product_id 
 GROUP BY 
 	p.category 
 HAVING  
 	sum(oi.quantity * oi.price_at_purchase) > 1;
 /* Задание 6 */
 SELECT 
 	country,
 	count(is_active) AS active_users_count 
 FROM 
 	users u 
 WHERE
 	is_active = TRUE 
 GROUP BY 
 	country
 HAVING 
 	count(is_active) > 2;
 /* Задание 7 */
 SELECT 
 	p.name AS product_name,
 	COALESCE (sum(oi.quantity), 0) AS total_quantity_sold 
 FROM
 	products p 
 LEFT JOIN
 	order_items oi ON p.id = oi.product_id 
 GROUP BY p.name
 ORDER BY 2 DESC 
 LIMIT 3
 /* Задание 8 
 * Условие для HAVING в задаче кривое, < 100 ничего нет
 * */
 SELECT 
 	order_id,
 	SUM(quantity * price_at_purchase) AS order_revenue 
 FROM	
 	order_items oi 
 GROUP BY
 	order_id
 HAVING 
 	SUM(quantity * price_at_purchase) < 10000
 ORDER BY 2;