SQL Exercise: Find customers who have placed no order or one or more

SQL JOINS: Exercise-11 with Solution

SQL statement to generate a report with customer name, city, order number, order date, order amount, salesperson name, and commission to determine if any of the existing customers have not placed orders or if they have placed orders through their salesman or by themselves.

Sample table: customer

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

Sample table: orders

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 table: salesman

 salesman_id |    name    |   city   | commission 
-------------+------------+----------+------------
        5001 | James Hoog | New York |       0.15
        5002 | Nail Knite | Paris    |       0.13
        5005 | Pit Alex   | London   |       0.11
        5006 | Mc Lyon    | Paris    |       0.14
        5007 | Paul Adam  | Rome     |       0.13
        5003 | Lauson Hen | San Jose |       0.12

Sample Solution:

-- Selecting specific columns and renaming one column for clarity
SELECT a.cust_name, a.city, b.ord_no,
       b.ord_date, b.purch_amt AS "Order Amount", 
       c.name, c.commission 
-- Specifying the tables to retrieve data from ('customer' as 'a', 'orders' as 'b', and 'salesman' as 'c')
FROM customer a 
-- Performing a left outer join based on the customer_id, including unmatched rows from 'customer'
LEFT OUTER JOIN orders b 
ON a.customer_id = b.customer_id 
-- Performing another left outer join with the result of the previous join and the 'salesman' table based on salesman_id
LEFT OUTER JOIN salesman c 
ON c.salesman_id = b.salesman_id;

Output of the Query:

cust_name	city		ord_no	ord_date	Order Amount	name		commission
Brad Guzan	London		70009	2012-09-10	270.65		Pit Alex	0.11
Nick Rimando	New York	70002	2012-10-05	65.26		James Hoog	0.15
Geoff Cameron	Berlin		70004	2012-08-17	110.50		Lauson Hen	0.12
Brad Davis	New York	70005	2012-07-27	2400.60		James Hoog	0.15
Nick Rimando	New York	70008	2012-09-10	5760.00		James Hoog	0.15
Fabian Johnson	Paris		70010	2012-10-10	1983.43		Mc Lyon		0.14
Geoff Cameron	Berlin		70003	2012-10-10	2480.40		Lauson Hen	0.12
Jozy Altidor	Moscow		70011	2012-08-17	75.29		Paul Adam	0.13
Nick Rimando	New York	70013	2012-04-25	3045.60		James Hoog	0.15
Graham Zusi	California	70001	2012-10-05	150.50		Nail Knite	0.13
Graham Zusi	California	70007	2012-09-10	948.50		Nail Knite	0.13
Julian Green	London		70012	2012-06-27	250.45		Nail Knite	0.13

Explanation:

The said SQL query is selecting the customer name, city, order number, order date, purchase amount, salesman name and commission from 3 tables customer aliased as a, orders aliased as b, and salesman aliased as c. It is joining these tables on the 'customer_id' and 'salesman_id' column respectively.
Additionally, it is using two LEFT OUTER JOINs, which will retrieve all records from the left table and the matching records from the right table. If no match is found on the right table, it will return NULL for the right table's fields.
This query will retrieve all customer details, order details and salesman details along with their commission even if some customer or salesman doesn't have any orders.

Visual Explanation:

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Query Visualization:

Duration:

Rows:

Cost:

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