oplossing How-to-code-subqueries.sql

How-to-code-subqueries.sql

@@ -1,19 +1,62 @@
 -- Exercise 1
-
+SELECT DISTINCT category_name
+FROM categories 
+WHERE category_id 
+IN (select product_id FROM products)
+ORDER BY category_name
+LIMIT 3;
 
 -- Exercise 2
-
+SELECT product_name, list_price
+FROM products
+GROUP BY product_name 
+HAVING AVG(list_price) > (SELECT AVG(list_price) FROM products)
+ORDER BY list_price DESC;
 
 -- Exercise 3
+SELECT category_name
+FROM categories c
+WHERE NOT exists
+(SELECT category_name, category_id 
+FROM products p
+WHERE c.category_id= p.category_id);
 
 
 -- Exercise 4
-
+SELECT email_address, os.order_id, SUM(item_price - discount_amount) * quantity AS 'order_total'
+FROM customers c
+JOIN orders os
+USING (customer_id)
+JOIN order_items oi
+USING (order_id)
+GROUP BY email_address, os.order_id;
 
 -- Exercise 4 Part 2
-
+SELECT 
+email_address,
+MAX(order_total)
+FROM (SELECT email_address, i.order_id, SUM((i.item_price - i.discount_amount) * quantity) AS 'order_total'
+	  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, order_id) f
+GROUP BY email_address;
 
 -- Exercise 5
-
+SELECT p1.product_name, p1.discount_percent
+FROM products p1
+WHERE p1.discount_percent NOT IN
+    (SELECT p2.discount_Percent
+     FROM Products p2
+     WHERE p1.product_name <> p2.product_name)
+ORDER BY product_name;
 
 -- Exercise 6
+SELECT email_address, order_id, order_date
+FROM customers c
+JOIN orders o
+ON c.customer_id = o.customer_id
+WHERE order_date = (SELECT MIN(order_date)
+					FROM orders
+					WHERE customer_id = o.customer_id);