Question 21080 in Mysql Question posted in
The official MySQL documentation can be found here.

SQL Consecutive Monthly Purchases – Mysql

I’m having great difficulty writing this query and cannot find any answers online which could be applied to my problem.

I have a couple of tables which looks similar to the below with. Each purchase date corresponds with an item purchased.

Cust_ID Purchase_Date
123 08/01/2022
123 08/20/2022
123 09/05/2022
123 10/08/2022
123 12/25/2022
123 01/26/2023

The result I am looking for should contain the customers ID, a range of the purchases, the number of consecutive months they had made a purchase (regardless of which day they purchased), and a count of how many purchases they had made in the time frame. The result should look something like the below for my example.

Cust_ID Min Purchase Date Max Purchase Date Consecutive Months No. Items Purchased
123 08/01/2022 10/08/2022 3 4
123 12/25/2022 01/26/2023 2 2

I have tried using CTEs with querys similar to

WITH CTE as

(
SELECT 
PaymentDate PD,
CustomerID CustID,
DATEADD(m, -ROW_NUMBER() OVER (PARTITION BY c.CustomerID ORDER BY 
DATEPART(m,PaymentDate)), PaymentDate) as TempCol1,
FROM customers as c
LEFT JOIN payments as p on c.customerid = p.customerid
GROUP BY c.CustomerID, p.PaymentDate
)

SELECT
CustID,
MIN(PD) AS MinPaymentDate,
MAX(PD) AS MaxPaymentDate,
COUNT(*) as ConsecutiveMonths,
FROM CTE
GROUP BY CustID, TempCol1

However, the above failed to properly count consecutive months. When the payment dates matched a month apart (e.g. 1/1/22 – 2/1/22), the query properly counts the consecutive months. However, if the dates do not match from month to month (e.g. 1/5/22 – 2/15/22), the count breaks.

Any guidance/help would be much appreciated!

Tags:

2

Answers


  1. 0

    You need to use the dense_rank function instead of the row_number, this will give the same rank for the same months and avoid breaking the grouping column. Also, you need to aggregate for ‘year-month’ of the grouping date column.

    with consecutive_months as
(
  select *,
     Purchase_Date - interval 
     dense_rank() over (partition by Cust_ID order by year(Purchase_Date), month(Purchase_Date)) 
     month as grp_date
  from payments
)
select C.Cust_ID, 
       min(M.Purchase_Date) as MinPurchaseDate,
       max(M.Purchase_Date) as MaxPurchaseDate,
       count(distinct year(M.Purchase_Date), month(M.Purchase_Date)) as ConsecutiveMonthsNo,
       count(M.Cust_ID) as ItemsPurchased 
from customers C left join consecutive_months M
on C.Cust_ID = M.Cust_ID
group by C.Cust_ID, year(M.grp_date), month(M.grp_date)

    See demo on MySQL

    You tagged your question with MySQL, while it seems that you posted an SQL Server query syntax, for SQL Server just use dateadd(month, -dense_rank() over (partition by Cust_ID order by year(Purchase_Date), month(Purchase_Date)), Purchase_Date).

    See demo on SQL Server.

    Login or Signup to reply.
  2. 0

    This is just a small enhancement on the answer already given by ahmed. If your date range for this query is more than a year, then year(M.Purchase_Date) + month(M.Purchase_Date) will be 2024 for both 2022-02-01 and 2023-01-01 as YEAR() and MONTH() both return integer values. This will return incorrect count of consecutive months. You can change this to use CONCAT() or FORMAT(). Also, the COUNT(*) for ItemsPurchased should be counting the right hand side of the join, as it is a LEFT JOIN.

    WITH consecutive_months AS
(
    SELECT *,
        DATEADD(
            month,
            -DENSE_RANK() OVER (
                PARTITION BY CustomerID
                ORDER BY YEAR(PaymentDate), MONTH(PaymentDate)
            ),
            PaymentDate
        ) AS grp_date
    FROM payments
)
SELECT
    C.CustomerID AS CustID,
    MIN(M.PaymentDate) AS MinPaymentDate,
    MAX(M.PaymentDate) AS MaxPaymentDate,
    COUNT(DISTINCT FORMAT(M.PaymentDate, 'yyyyMM')) AS ConsecutiveMonths,
    COUNT(M.CustomerID) AS ItemsPurchased 
FROM customers C
LEFT JOIN consecutive_months M
    ON C.CustomerID = M.CustomerID
GROUP BY C.CustomerID, YEAR(M.grp_date), MONTH(M.grp_date)

    Here’s a db<>fiddle

    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search