SQL Exercises: Customer order amounts that are on or above average

SQL SUBQUERY : Exercise-13 with Solution

13. From the following tables write a SQL query to find those orders that are equal or higher than the average amount of the orders. Return ord_no, purch_amt, ord_date, customer_id and salesman_id.

ord_no      purch_amt   ord_date    customer_id  salesman_id
----------  ----------  ----------  -----------  -----------
70001       150.5       2012-10-05  3005         5002
70009       270.65      2012-09-10  3001         5005
70002       65.26       2012-10-05  3002         5001
70004       110.5       2012-08-17  3009         5003
70007       948.5       2012-09-10  3005         5002
70005       2400.6      2012-07-27  3007         5001
70008       5760        2012-09-10  3002         5001
70010       1983.43     2012-10-10  3004         5006
70003       2480.4      2012-10-10  3009         5003
70012       250.45      2012-06-27  3008         5002
70011       75.29       2012-08-17  3003         5007
70013       3045.6      2012-04-25  3002         5001

Sample Solution:

-- Selecting all columns from the 'orders' table (aliased as 'a')
SELECT * 
-- Specifying the table to retrieve data from ('orders' as 'a')
FROM orders a
-- Filtering the results based on the condition that 'purch_amt' is greater than or equal to the average 'purch_amt' for the same 'customer_id'
WHERE purch_amt >=
    -- Subquery: Calculating the average 'purch_amt' from the 'orders' table (aliased as 'b') for the same 'customer_id'
    (SELECT AVG(purch_amt) FROM orders b 
     WHERE b.customer_id = a.customer_id);

Output of the Query:

ord_no	purch_amt	ord_date	customer_id	salesman_id
70009	270.65		2012-09-10	3001		5005
70005	2400.60		2012-07-27	3007		5001
70008	5760.00		2012-09-10	3002		5001
70010	1983.43		2012-10-10	3004		5006
70003	2480.40		2012-10-10	3009		5003
70011	75.29		2012-08-17	3003		5007
70013	3045.60		2012-04-25	3002		5001
70007	948.50		2012-09-10	3005		5002
70012	250.45		2012-06-27	3008		5002

Explanation:

The said SQL query that selects all columns from the 'orders' table (aliased as 'a') where the "purch_amt" column is greater than or equal to the average "purch_amt" value for the corresponding "customer_id" in the same 'orders' table (aliased as 'b'). In other words, it's retrieving all the rows from 'orders' table where the "purch_amt" is greater than or equal to the average purchase amount made by the customer.

Visual Explanation:

Practice Online

Sample Database: inventory

Query Visualization:

Duration:

Rows:

Cost:

Contribute your code and comments through Disqus.

Previous SQL Exercise: Find all orders with above-average amounts.
Next SQL Exercise: Sums of all order amounts, grouped by date.