w3resource

SQL Exercise: Salesmen who works either for one or more customer

SQL JOINS: Exercise-13 with Solution

From the following tables write a SQL query to list all salespersons along with customer name, city, grade, order number, date, and amount. Condition for selecting list of salesmen : 1. Salesmen who works for one or more customer or, 2. Salesmen who not yet join under any customer, Condition for selecting list of customer : 3. placed one or more orders, or 4. no order placed to their salesman.

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: 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 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 Solution:

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

Output of the Query:

cust_name	city		grade	Salesman	ord_no	ord_date	purch_amt
Brad Guzan	London			Pit Alex	70009	2012-09-10	270.65
Nick Rimando	New York	100	James Hoog	70002	2012-10-05	65.26
Geoff Cameron	Berlin		100	Lauson Hen	70004	2012-08-17	110.50
Brad Davis	New York	200	James Hoog	70005	2012-07-27	2400.60
Nick Rimando	New York	100	James Hoog	70008	2012-09-10	5760.00
Fabian Johnson	Paris		300	Mc Lyon		70010	2012-10-10	1983.43
Geoff Cameron	Berlin		100	Lauson Hen	70003	2012-10-10	2480.40
Jozy Altidor	Moscow		200	Paul Adam	70011	2012-08-17	75.29
Nick Rimando	New York	100	James Hoog	70013	2012-04-25	3045.60
Graham Zusi	California	200	Nail Knite	70001	2012-10-05	150.50
Graham Zusi	California	200	Nail Knite	70007	2012-09-10	948.50
Julian Green	London		300	Nail Knite	70012	2012-06-27	250.45

Explanation:

The said SQL query is performing a right outer join on three tables: the customer table alias a, the salesman table alias b, and the orders table alias c.
First, it performs a right outer join on the customer table and the salesman table using the 'salesman_id' column.
Then it performs another right outer join on the result of the first join and the orders table using the 'customer_id' column.
It is then selecting the 'cust_name', 'city', 'grade', 'name' as 'Salesman', 'ord_no', 'ord_date', and 'purch_amt' columns from the three tables.
This query will select all the rows from orders table and any matching rows from customer table and salesman table. If there is no match, it will return NULL for the non-matching columns of customer table and salesman table.

Visual Explanation:

Result of a list for the salesmen who works either for one or more customer or not yet join under any of the customer who placed either one or more orders or no order to their supplier
Result of a list for the salesmen who works either for one or more customer or not yet join under any of the customer who placed either one or more orders or no order to their supplier

Practice Online


Query Visualization:

Duration:

Query visualization of Display the list for the salesmen who works either for one or more customer or not yet join under any of the customers who placed either one or more orders or no order to their supplier - Duration

Rows:

Query visualization of Display the list for the salesmen who works either for one or more customer or not yet join under any of the customers who placed either one or more orders or no order to their supplier - Rows

Cost:

Query visualization of Display the list for the salesmen who works either for one or more customer or not yet join under any of the customers who placed either one or more orders or no order to their supplier - Cost

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous SQL Exercise: Salesmen works either for one or more customer or not.
Next SQL Exercise: Salesmen work for one or more customers or yet to join.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.



Follow us on Facebook and Twitter for latest update.