SQL Exercises: Orders higher than any amount for a London customer

SQL SUBQUERY: Exercise-25 with Solution

25. From the following tables write a SQL query to find those orders where every order amount is less than the maximum order amount of a customer who lives in London City. 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

customer_id  cust_name     city        grade       salesman_id
-----------  ------------  ----------  ----------  -----------
3002         Nick Rimando  New York    100         5001
3005         Graham Zusi   California  200         5002
3001         Brad Guzan    London      100         5005
3004         Fabian Johns  Paris       300         5006
3007         Brad Davis    New York    200         5001
3009         Geoff Camero  Berlin      100         5003
3008         Julian Green  London      300         5002
3003         Jozy Altidor  Moncow      200         5007

Sample Solution:

-- Selecting all columns from the 'orders' table
SELECT *
-- Specifying the table to retrieve data from ('orders')
FROM orders
-- Filtering the results based on the condition that 'purch_amt' is less than the maximum 'purch_amt' value returned by a correlated subquery
WHERE purch_amt < 
   -- Correlated Subquery: Selecting the maximum 'purch_amt' value from the 'orders' table (aliased as 'a') where 'customer_id' matches the outer query's 'customer_id' and 'city' is 'London' in the 'customer' table (aliased as 'b')
   (SELECT MAX(purch_amt)
	FROM orders a, customer b
	WHERE  a.customer_id = b.customer_id
	AND b.city = 'London');

Output of the Query:

ord_no	purch_amt	ord_date	customer_id	salesman_id
70002	65.26		2012-10-05	3002		5001
70004	110.50		2012-08-17	3009		5003
70011	75.29		2012-08-17	3003		5007
70001	150.50		2012-10-05	3005		5002
70012	250.45		2012-06-27	3008		5002

Explanation:

The above SQL query is selecting all columns (*) from the 'orders' table where the "purch_amt" value of the row is less than the maximum "purch_amt" value returned by the subquery.
The subquery is joining the 'orders' table (alias as 'a') with the 'customer' table (alias as 'b') on the "customer_id" column and then using the MAX function to select the highest "purch_amt" value from the 'orders' table for the rows where the "city" value of the 'customer' table is 'London'. The outer query will only return the rows from the "orders" table whose purch_amt is less than the maximum purch_amt of orders placed by the customer from London.

Visual Explanation:

Practice Online

Sample Database: inventory

Query Visualization:

Duration:

Rows:

Cost:

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