oplossing How-to-code-summary-queries.sql

How-to-code-summary-queries.sql

@@ -1,19 +1,51 @@
 -- Exercise 1
-
+SELECT COUNT(order_id), SUM(tax_amount)
+FROM orders;
 
 -- Exercise 2
-
+SELECT category_name, COUNT(product_id) AS count, MAX(list_price) AS max_price
+FROM categories c JOIN products p
+ON c.category_id = p.category_id
+GROUP BY category_name
+ORDER BY category_name DESC;
 
 -- Exercise 3
-
+SELECT email_address, SUM(item_price) * quantity AS item_price_total ,
+       SUM(discount_amount) * quantity AS discount_amount_total
+FROM orders o JOIN order_items i
+ON o.order_id = i.order_id
+JOIN customers c ON o.customer_id = c.customer_id
+GROUP BY c.customer_id
+ORDER BY SUM(item_price) * quantity DESC;
 
 -- Exercise 4
-
+SELECT c.email_address,	COUNT(oi.order_id) AS 'order_count',
+	   SUM(item_price - discount_amount) * quantity AS 'order_total'
+FROM orders ord JOIN order_items oi ON o.order_id = oi.order_id
+JOIN customers c ON o.customer_id = c.customer_id
+GROUP BY email_address
+HAVING COUNT(oi.order_id) > 1
+ORDER BY SUM(item_price - discount_amount) * COUNT(i.order_id) DESC;
 
 -- Exercise 5
+SELECT email_address, COUNT(o.order_id) AS 'order_count',
+       SUM(item_price - discount_amount) * quantity AS 'order_total'
+FROM orders o JOIN order_items oi ON o.order_id = oi.order_id
+              JOIN customers c ON o.customer_id = c.customer_id
+WHERE oi.item_price > 400
+GROUP BY email_address
+HAVING COUNT(oi.order_id) > 1
+ORDER BY SUM(item_price - discount_amount) * COUNT(oi.order_id)  DESC;
 
 
 -- Exercise 6
-
+SELECT product_name, SUM(item_price - discount_amount) * quantity AS 'total_amount'
+FROM products p JOIN order_items i ON p.product_id = i.product_id
+GROUP BY product_name WITH ROLLUP;
 
 -- Exercise 7
+SELECT email_address, COUNT(product_id) AS 'distinct_products'
+FROM customers c JOIN orders o ON c.customer_id = o.customer_id
+				 JOIN order_items i ON o.order_id = i.order_id
+GROUP BY email_address
+HAVING COUNT(o.order_id) > 1;