SQL Filter rows on date difference

Question

MySQL 5.6 Have a table which I query:

SELECT DISTINCT client, date_buy
FROM tickets
ORDER BY client, date_buy

Which gives me almost the result I need:

------------------------------
client          | date_buy
-------------------------------
0027847524333  | 2018-06-13 16:03:43
0027847524333  | 2018-06-13 16:13:01
0027847524333  | 2018-06-18 22:03:01
0033652356025  | 2018-06-16 17:22:56
00353857861869 | 2018-08-13 17:37:56

What do I need to add to query so I could leave only clients where it has been more than 24h between orders?

Desired result:

------------------------------
client          | date_buy
-------------------------------
0027847524333  | 2018-06-13 16:03:43
0027847524333  | 2018-06-18 22:03:01
0033652356025  | 2018-06-16 17:22:56
00353857861869 | 2018-08-13 17:37:56

Update: I would like to exclude repetitive sales for one customer that happened less than 24 hours each other.

Answer 1

You can try to use group by with DATE(date_buy) and client instead of DISTINCT . Then get MIN(date_buy) be your date_buy

Schema (MySQL v5.7)

CREATE TABLE tickets(
  client varchar(50),
  date_buy datetime
);


insert into tickets values ('0027847524333', '2018-06-13 16:03:43');
insert into tickets values ('0027847524333', '2018-06-13 16:13:01');
insert into tickets values ('0027847524333', '2018-06-18 22:03:01');
insert into tickets values ('0033652356025', '2018-06-16 17:22:56');
insert into tickets values ('00353857861869','2018-08-13 17:37:56');

---

Query #1

SELECT client, MIN(date_buy) date_buy
FROM tickets
GROUP BY DATE(date_buy),client
ORDER BY client, MIN(date_buy);

| client         | date_buy            |
| -------------- | ------------------- |
| 0027847524333  | 2018-06-13 16:03:43 |
| 0027847524333  | 2018-06-18 22:03:01 |
| 0033652356025  | 2018-06-16 17:22:56 |
| 00353857861869 | 2018-08-13 17:37:56 |

---

View on DB Fiddle

Answer 2

you can do this :

    SELECT * 
FROM tickets t_1, tickets t_2
where t_1.client = t_2.client
and DATEDIFF(hour, t_1.date_buy, t_2.date_buy) < 24

Answer 3

It seems that the date_buy column is of smalldatetime type (if not, you can easily parse it to be).

Here's one way you could implement the query. Start by creating a table of all later clients where it has been less than or equal to 24 hours since a previous order:

CREATE TABLE helper AS
SELECT DISTINCT b.client AS cl
FROM tickets AS a, tickets AS b
WHERE DATEDIFF(hour, a.date_buy, b.date_buy) <= 24 AND a.client = b.client

Then, select all other clients from tickets :

SELECT DISTINCT client, date_buy
FROM tickets, helper
WHERE client <> cl
ORDER BY client, date_buy

You can obtain the same result by deleting entries from tickets , though this would mutate the original table. This solution takes into account hour differences across the day change .

Answer 4

Here is a solution using LAG

SELECT client, date_buy
FROM (SELECT client, date_buy, 
        DATE_ADD(IFNULL(LAG(date_buy, 1) OVER(PARTITION BY client ORDER BY client, date_buy ASC), '1900-01-01 00:00:00'), INTERVAL 24 HOUR) prev_date
      FROM tickets) sub
WHERE date_buy > prev_date