显示客户每天的花费以及他们是否在前一天花费过 (SQL)

Question

I am trying to create a new row for each customer who spends per day, and a column that indicates whether they spent money the day before or not. If a customer spends twice a day, they'd still only have 1 row in the table. If the customer spent money the previous day, then it would show up as TRUE.

This is the original table below:

+---------------------+-------------+-----------------+
| datetime            | customer_id | amount          |
+---------------------+-------------+-----------------+
| 2018-03-01 03:00:00 | 3786        | 14.00000        |
| 2018-03-02 17:00:00 | 5678        | 25.00000        |
| 2018-07-09 18:00:00 | 5647        | 1000.99000      |
| 2018-08-17 19:00:00 | 5267        | 45.00000        |
| 2018-08-25 08:00:00 | 3456        | 78.00000        |
| 2018-08-25 17:00:00 | 3456        | 25.00000        |
| 2018-08-26 03:00:00 | 3456        | 34.90000        |
| 2019-02-03 08:00:00 | 3468        | 0.00000         |
| 2019-03-09 06:00:00 | 1111        | 100.00000       |
| 2019-05-25 14:00:00 | 3456        | 15.00000        |
| 2019-07-02 14:00:00 | 88889       | 45.00000        |
| 2019-07-04 03:00:00 | 8979        | 9.00000         |
| 2019-07-09 14:00:00 | 4567        | 9.99000         |
| 2019-08-25 08:00:00 | 1234        | 88.00000        |
| 2019-08-30 09:31:00 | 1234        | 30.00000        |
| 2019-08-30 12:00:00 | 9876        | 55.00000        |
| 2019-09-01 13:00:00 | 88889       | 23.00000        |
+---------------------+-------------+-----------------+

This is the CREATE statement:

 CREATE TABLE IF NOT EXISTS `spend` ( `datetime` datetime NOT NULL, `customer_id` int(11) NOT NULL, `amount` decimal(10, 5) NOT NULL, PRIMARY KEY (`datetime`)) DEFAULT CHARSET=utf8mb4; INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-03-01 03:00:00', 3786, 14.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-03-02 17:00:00', 5678, 25.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-07-09 18:00:00', 5647, 1000.99000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-08-17 19:00:00', 5267, 45.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-08-25 08:00:00', 3456, 78.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-08-25 17:00:00', 3456, 25.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-08-26 03:00:00', 3456, 34.90000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-02-03 08:00:00', 3468, 0.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-03-09 06:00:00', 1111, 100.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-05-25 14:00:00', 3456, 15.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-07-02 14:00:00', 88889, 45.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-07-04 03:00:00', 8979, 9.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-07-09 14:00:00', 4567, 9.99000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-08-25 08:00:00', 1234, 88.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-08-30 09:31:00', 1234, 30.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-08-30 12:00:00', 9876, 55.00000); INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-09-01 13:00:00', 88889, 23.00000);

Here's what I've got so far:

 SELECT CAST(datetime AS DATE) AS day, COUNT(DISTINCT customer_id) AS daily_spend, FROM spend WHERE amount is not NULL ORDER BY date;

This code is not working at the moment, but I'm trying to fix it as best I can.

I've had a look through some of the posts, but the closest I could find was this: count transaction per day

I'm trying to produce a table that looks like this:

+------------+-------------+--------------------+
| day        | customer_id | spent_previous_day |
+------------+-------------+--------------------+
| 2018-03-01 | 3786        | FALSE              |
+------------+-------------+--------------------+
| 2018-03-02 | 5678        | FALSE              |
+------------+-------------+--------------------+
| 2018-07-09 | 5647        | FALSE              |
+------------+-------------+--------------------+
| 2018-08-17 | 5267        | FALSE              |
+------------+-------------+--------------------+
| 2018-08-25 | 3456        | FALSE              |
+------------+-------------+--------------------+
| 2018-08-26 | 3456        | TRUE               |
+------------+-------------+--------------------+
| 2019-02-03 | 3468        | FALSE              |
+------------+-------------+--------------------+
| 2019-03-09 | 1111        | FALSE              |
+------------+-------------+--------------------+
| 2019-05-25 | 3456        | FALSE              |
+------------+-------------+--------------------+
| 2019-07-02 | 88889       | FALSE              |
+------------+-------------+--------------------+
| 2019-07-04 | 8979        | FALSE              |
+------------+-------------+--------------------+
| 2019-07-09 | 4567        | FALSE              |
+------------+-------------+--------------------+
| 2019-08-25 | 1234        | FALSE              |
+------------+-------------+--------------------+
| 2019-08-30 | 1234        | FALSE              |
+------------+-------------+--------------------+
| 2019-08-30 | 9876        | FALSE              |
+------------+-------------+--------------------+
| 2019-09-01 | 88889       | FALSE              |
+------------+-------------+--------------------+

Edit: This is the current code I'm using, based on the advice I've received.

 select customer_id, CAST(datetime AS DATE) AS day, max(date(datetime)) over (partition by customer_id order by CAST(datetime AS DATE) range between interval 1 day preceding and interval 1 day preceding ) is not null AS spent_previous_day from spend

This is the resulting table:

+------------+-------------+--------------------+
| day        | customer_id | spent_previous_day |
+------------+-------------+--------------------+
| 2019-03-09 | 1111        | 0                  |
+------------+-------------+--------------------+
| 2019-08-25 | 1234        | 0                  |
+------------+-------------+--------------------+
| 2019-08-30 | 1234        | 0                  |
+------------+-------------+--------------------+
| 2018-08-25 | 3456        | 0                  |
+------------+-------------+--------------------+
| 2018-08-25 | 3456        | 0                  |
+------------+-------------+--------------------+
| 2018-08-26 | 3456        | 1                  |
+------------+-------------+--------------------+
| 2019-05-25 | 3456        | 0                  |
+------------+-------------+--------------------+
| 2019-02-03 | 3468        | 0                  |
+------------+-------------+--------------------+

I've tried to do a GROUP BY day, customer_id , but it comes up as an error.

Answer 1

Assuming that customers do not make multiple purchases on the same day, just use lag() :

select t.*,
       ( date(lag(datetime) over (partition by customer_id order by datetime)) = date(datetime) - interval 1 day
       ) as prev_day_flag
from spend t;

If you can have duplicates, then try this instead of lag() :

max(date(datetime)) over (partition by customer_id
                          order by date(datetime) 
                          range between interval 1 day preceding and interval 1 day preceding
                         ) is not null

EDIT:

If you want one row per customer per day:

select s.*,
       ( date(lag(dte) over (partition by customer_id order by dte)) = dte - interval 1 day
       ) as prev_day_flag
from (select customer_id, date(datetime) as dte, sum(amount) as amount
      from spend s
      group by customer_id, date(datetime)
     ) s;