确定每月第一次在公司花钱的客户（mySQL）

Question

我有一个下表，我正在尝试确定每个月第一次在公司花钱的用户数量。

我想要一个结果表，其中包含新用户、月份和年份作为列。

在人们对这篇文章投反对票之前，我已经浏览了各种帖子，似乎找不到类似的方法来解决这个问题。我在下面包含的代码是基于我从相关帖子中拼凑出来的。

这是原始表格：

+---------------------+-------------+-----------------+
| datetime            | customer_id | amount          |
+---------------------+-------------+-----------------+
| 2018-03-01 03:00:00 | 3786        | 14              |
+---------------------+-------------+-----------------+
| 2018-03-02 17:00:00 | 5678        | 25              |
+---------------------+-------------+-----------------+
| 2018-08-17 19:00:00 | 5267        | 45              |
+---------------------+-------------+-----------------+
| 2018-08-25 08:00:00 | 3456        | 78              |
+---------------------+-------------+-----------------+
| 2018-08-25 17:00:00 | 3456        | 25              |
+---------------------+-------------+-----------------+
| 2019-05-25 14:00:00 | 3456        | 15              |
+---------------------+-------------+-----------------+
| 2019-07-02 14:00:00 | 88889       | 45              |
+---------------------+-------------+-----------------+
| 2019-08-25 08:00:00 | 1234        | 88              |
+---------------------+-------------+-----------------+
| 2019-08-30 09:31:00 | 1234        | 30              |
+---------------------+-------------+-----------------+
| 2019-08-30 12:00:00 | 9876        | 55              |
+---------------------+-------------+-----------------+
| 2019-09-01 13:00:00 | 88889       | 23              |
+---------------------+-------------+-----------------+

这是 CREATE 语句：

CREATE TABLE IF NOT EXISTS `spend` ( `datetime` datetime NOT NULL, `customer_id` int(11) NOT NULL, `amount` int(11) NOT NULL, PRIMARY KEY (`datetime`)) DEFAULT CHARSET=utf8mb4;
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-03-01 03:00:00', 3786, 14);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-03-02 17:00:00', 5678, 25);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-08-17 19:00:00', 5267, 45);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-08-25 08:00:00', 3456, 78);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2018-08-25 17:00:00', 3456, 25);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-05-25 14:00:00', 3456, 15);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-07-02 14:00:00', 88889, 45);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-08-25 08:00:00', 1234, 88);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-08-30 09:31:00', 1234, 30);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-08-30 12:00:00', 9876, 55);
INSERT INTO `spend` (`datetime`, `customer_id`, `amount`) VALUES ('2019-09-01 13:00:00', 88889, 23);

这是我想出的代码：

SELECT S.datetime, S.customer_id, S.amount 
FROM spend S
INNER JOIN
    (SELECT customer_id, MIN(datetime) AS first_occurence
    FROM spend
    GROUP BY customer_id) X
ON S.customer_id = X.customer_id AND S.datetime = X.first_occurence

这是结果表：

+------------------+-------------+-------+
| datetime         | customer_id |amount |
+------------------+-------------+-------+
| 01/03/2018 03:00 | 3786        | 14    |
+------------------+-------------+-------+
| 02/03/2018 17:00 | 5678        | 25    |
+------------------+-------------+-------+
| 17/08/2018 19:00 | 5267        | 45    |
+------------------+-------------+-------+
| 25/08/2018 08:00 | 3456        | 78    |
+------------------+-------------+-------+
| 02/07/2019 14:00 | 88889       | 45    |
+------------------+-------------+-------+
| 25/08/2019 08:00 | 1234        | 88    |
+------------------+-------------+-------+
| 30/08/2019 12:00 | 9876        | 55    |
+------------------+-------------+-------+

这是表格应该是什么样子的示例：

+-----------+-------+------+
| new_users | month | year |
+-----------+-------+------+
| 2         | 3     | 2018 |
+-----------+-------+------+
| 3         | 8     | 2018 |
+-----------+-------+------+
| 1         | 5     | 2019 |
+-----------+-------+------+
| 1         | 7     | 2019 |
+-----------+-------+------+
| 3         | 8     | 2019 |
+-----------+-------+------+
| 1         | 9     | 2019 |
+-----------+-------+------+

Answer 1

你的开始是正确的。现在将其用作子查询以按月获取计数。

SELECT COUNT(*) AS new_users, MONTH(datetime) AS month, YEAR(datetime) AS year
FROM (
    SELECT S.datetime, S.customer_id, S.amount 
    FROM spend S
    INNER JOIN
        (SELECT customer_id, MIN(datetime) AS first_occurence
        FROM spend
        GROUP BY customer_id) X
    ON S.customer_id = X.customer_id AND S.datetime = X.first_occurence
) AS x
GROUP BY month, year
ORDER BY year, month

实际上，您甚至不需要子查询中的联接，因为您没有在最终结果中使用第一次购买的金额。

SELECT COUNT(*) AS new_users, MONTH(datetime) AS month, YEAR(datetime) AS year
FROM (
    SELECT customer_id, MIN(datetime) AS datetime
    FROM spend
    GROUP BY customer_id
) AS x
GROUP BY month, year
ORDER BY year, month

Answer 2

您不需要两级深度子查询。您可以简单地找到客户第一次使用MIN() 花钱的时间，然后从最小日期时间值中提取YEAR() 和MONTH() 来计算用户数量：

SELECT 
  YEAR(min_dt) y,
  MONTH(min_dt) m,
  COUNT(*) AS new_customers 
FROM 
(
  SELECT customer_id, MIN(datetime) AS min_dt 
  FROM spend 
  GROUP BY customer_id 
) t
GROUP BY y, m

结果

| y    | m   | new_customers |
| ---- | --- | ------------- |
| 2018 | 3   | 2             |
| 2018 | 8   | 2             |
| 2019 | 7   | 1             |
| 2019 | 8   | 2             |

---

View on DB Fiddle

Answer 3

使用 ROW_NUMBER() 窗口函数：

select 
  count(*) new_users,
  month(t.datetime) month,
  year(t.datetime) year
from (
  select *,
    row_number() over (partition by customer_id order by datetime) rn
  from spend
) t
where t.rn = 1
group by year, month
order by year, month

请参阅demo 获取您的示例数据。
结果：

| new_users | month | year |
| --------- | ----- | ---- |
| 2         | 3     | 2018 |
| 2         | 8     | 2018 |
| 1         | 7     | 2019 |
| 2         | 8     | 2019 |

Answer 4

你也可以

select   
count(*) new_users,
month(datetime) month,
year(datetime) year
from spend
where datetime in (select min(datetime) from spend group by customer_id)
group by year, month
order by year, month;