SQL: How to find top customers that pay 80% of revenue? - mysql

Question

Let's say I have a table TRANSACTIONS:
desc customer_transactions;
+------------------------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------------------------+--------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| transactionID | varchar(128) | YES | | NULL | |
| customerID | varchar(128) | YES | | NULL | |
| amountAuthorized | DECIMAL(5,2) | YES | | NULL | |
| createdDatetime | datetime | YES | | NULL | |
+------------------------------+--------------+------+-----+---------+----------------+
This table has records of credit card transactions for a SAAS business for the last 5 years. The business has a typical monthly subscription model, where customers automatically charged based on their plan.
I need to find the top customers that are responsible for 80% of all revenue (per time period). The SAAS business is very uneven, because some customers pay 10/month, others may pay in thousands per month.
I will add a "time period" filter later, just need help with aggregation.
I want to generate a report where I only select the customers that generated 80% of revenue in this format:
+------------+-------+
| customerID | Total |
+------------+-------+
Not sure why this question was "on hold". I just need help writing a query and do not have enough experience with SQL. Basically, the title of the question states what is needed here:
I need to list customers and their corresponding totals, however, only need to select those customers that make up 80% of total revenue. The report needs to aggregate a total per customer.
Using MariaDB version 10.3.9

Answer 1

This is the kind of thing you need to use window functions for.
WITH
-- define some sample data,
-- where the sum total of amountAuthorized is 10,000
customer_transactions( `id`, transactionID, customerID,
amountAuthorized, createdDatetime) AS
(
SELECT 1, 1, 1, 5000, '2018-08-01'
UNION ALL SELECT 2, 2, 2, 2000, '2018-08-01'
UNION ALL SELECT 3, 3, 3, 1000, '2018-08-01'
UNION ALL SELECT 4, 4, 4, 1000, '2018-08-01'
UNION ALL SELECT 5, 5, 5, 1000, '2018-08-01'
)
-- a query that gives us the running total, sorted to give us the biggest customers first.
-- note that the additional sorts affect what customers might be returned.
,running_totals AS
(
SELECT *, SUM(amountAuthorized) OVER (ORDER BY amountAuthorized DESC, createdDatetime DESC, `id`) AS runningTotal
FROM customer_transactions
)
SELECT *
FROM running_totals
WHERE runningTotal <= ( SELECT 0.8 * SUM(amountAuthorized)
FROM customer_transactions)
Note that this takes into account (no pun intended) all data in the table. When you want to only look at a specific time period, you might want to create an intermediate CTE that filters out the dates you want.

Answer 2

You will find that surprisingly close 20% of the customers account for the 80%. See the 80/20 rule .
But, if you don't want to go that direction, you have 2 options:
Switch to MySQL 8.0 or MariaDB 10.1 in order to use 'windowing' functions; or
Use #variables to produce a running total, then (in an outer query) grab the desired rows.
Since you are using MariaDB 10.3.9, the windowing seems to be the way to go. But first, you need a separate query (or derived table) that computes the total revenue so you can get 80% of it.
Suggest
SELECT #revenue80 := 0.8 * SUM(amountAuthorized)
FROM customer_transactions
Then use #revenue80 inside the WHERE that Zack suggests.
I see that each amount can be no more than 999.99. Really? Is this a coffee shop?

Answer 3

Use the following:
SELECT
ct1.customerID,
SUM(ct1.amountAuthorized) as Total,
100 * (SUM(ct1.amountAuthorized) / ct3.total_revenue) as percent_revenue
FROM
customer_transactions ct1
CROSS JOIN (SELECT SUM(amountAuthorized) AS total_revenue
FROM customer_transactions ct2) AS ct3
GROUP BY
ct1.customerID
HAVING percent_revenue >= 80