w3resource

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


Sample table: orders


Sample table: salesman


Sample Solution:

SELECT a.cust_name,a.city, b.ord_no,
b.ord_date,b.purch_amt AS "Order Amount", 
c.name,c.commission 
FROM customer a 
LEFT OUTER JOIN orders b 
ON a.customer_id=b.customer_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:

Result of a report with customer name, city, order number, order date, order amount salesman name and commission
Result of a report with customer name, city, order number, order date, order amount salesman name and commission

Practice Online


Query Visualization:

Duration:

Query visualization of Make a report with customer name, city, order number, date, amount salesman name and commission to find either any of the existing customers have placed no order or placed one or more orders by their salesman or by own - Duration

Rows:

Query visualization of Make a report with customer name, city, order number, date, amount salesman name and commission to find either any of the existing customers have placed no order or placed one or more orders by their salesman or by own - Rows

Cost:

Query visualization of Make a report with customer name, city, order number, date, amount salesman name and commission to find either any of the existing customers have placed no order or placed one or more orders by their salesman or by own - Cost

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

Previous SQL Exercise: Customer have placed no order or one or more orders.
Next SQL Exercise: Salesmen works either for one or more customer or not.

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.

SQL: Tips of the Day

Grouped LIMIT in PostgreSQL: Show the first N rows for each group?

db=# SELECT * FROM xxx;
 id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  3 |          1 | C
  4 |          1 | D
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
  8 |          2 | H
(8 rows)

I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:

id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
(5 rows)

PostgreSQL v9.3 you can do a lateral join

select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
    select * from t t_inner
    where t_inner.section_id = t_outer.section_id
    order by t_inner.name
    limit 2
) t_top on true
order by t_outer.section_id;

Database: PostgreSQL

Ref: https://bit.ly/3AfYwZI

 





We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook