首次下订单时获取日期和唯一客户数

Question

我有一个名为 orders 的表，如下所示：

+--------------+---------+------+-----+---------+-------+
| Field        | Type    | Null | Key | Default | Extra |
+--------------+---------+------+-----+---------+-------+
| id           | int(11) | YES  |     | NULL    |       |
| memberid     | int(11) | YES  |     | NULL    |       |
| deliverydate | date    | YES  |     | NULL    |       |
+--------------+---------+------+-----+---------+-------+

其中包含以下数据：

+------+----------+--------------+
| id   | memberid | deliverydate |
+------+----------+--------------+
|    1 |      991 | 2019-10-25   |
|    2 |      991 | 2019-10-26   |
|    3 |      992 | 2019-10-25   |
|    4 |      992 | 2019-10-25   |
|    5 |      993 | 2019-10-24   |
|    7 |      994 | 2019-10-21   |
|    6 |      994 | 2019-10-26   |
|    8 |      995 | 2019-10-26   |
+------+----------+--------------+

我想要一个返回每个唯一日期的结果集，以及一个单独的列，显示当天有多少客户下了他们的第一个订单。

我在以正确方式查询时遇到问题，尤其是当数据包含同一客户同一天的多个订单时。

我的方法是

1. 获取在我要查看的时间段内下订单的所有唯一会员 ID
2. 通过比较该时间段前下单的memberid，过滤掉该时间段内首次下单的会员
3. 按交货日期分组，并计算所有唯一的 memberid（但这显然每天单独计算唯一的 memberid！）

下面是对应的SQL：

SELECT deliverydate,COUNT(DISTINCT memberid) FROM orders
WHERE 
    MemberId IN (SELECT DISTINCT memberid FROM orders WHERE deliverydate BETWEEN '2019-10-25' AND '2019-10-26')
AND NOT 
    MemberId In (SELECT DISTINCT memberid FROM orders WHERE deliverydate < '2019-10-25')
GROUP BY deliverydate
ORDER BY deliverydate ASC;

但这会导致上述数据如下：

+--------------+--------------------------+
| deliverydate | COUNT(DISTINCT memberid) |
+--------------+--------------------------+
| 2019-10-25   |                        2 |
| 2019-10-26   |                        2 |
+--------------+--------------------------+

2019-10-26 的计数应该是 1。

感谢任何帮助:)

Answer 1

你可以聚合两次：

select first_deliverydate, count(*) cnt
from (
    select min(deliverydate) first_deliverydate
    from orders
    group by memberid
) t
group by first_deliverydate
order by first_deliverydate

子查询为您提供每个成员的第一个订单数据，然后外部查询按第一个订单日期聚合和计数。

此demo on DB Fiddle 与您的示例数据返回：

first_deliverydate | cnt :----------------- | --: 2019-10-21 | 1 2019-10-24 | 1 2019-10-25 | 2 2019-10-26 | 1

---

在 MySQL 8.0 中，这也可以通过窗口函数来实现：

select deliverydate first_deliverydate, count(*) cnt
from (
    select deliverydate, row_number() over(partition by memberid order by deliverydate) rn
    from orders
) t
where rn = 1
group by deliverydate
order by deliverydate

Demo on DB Fiddle

Answer 2

您必须首先弄清楚第一个交货日期是什么时候：

SELECT firstdeliverydate,COUNT(DISTINCT memberid) FROM (
select memberid, min(deliverydate) as firstdeliverydate
from orders
WHERE 
    MemberId IN (SELECT DISTINCT memberid FROM orders WHERE deliverydate BETWEEN '2019-10-25' AND '2019-10-26')
AND NOT 
    MemberId In (SELECT DISTINCT memberid FROM orders WHERE deliverydate < '2019-10-25')
group by memberid)
t1
group by firstdeliverydate

Answer 3

使用NOT EXISTS 获取每个客户的第一个订单，然后使用GROUP BY deliverydate 计算下订单的不同客户：

select o.deliverydate, count(distinct o.memberid) counter
from orders o
where not exists (
  select 1 from orders
  where memberid = o.memberid and deliverydate < o.deliverydate
)  
group by o.deliverydate

请参阅demo。
结果：

| deliverydate        | counter |
| ------------------- | ------- |
| 2019-10-21 00:00:00 | 1       |
| 2019-10-24 00:00:00 | 1       |
| 2019-10-25 00:00:00 | 2       |
| 2019-10-26 00:00:00 | 1       |

但是，如果您想要表格中所有日期的结果，包括那些没有新客户订单的日期（所以计数器将是 0）：

select d.deliverydate, count(distinct o.memberid) counter
from (
  select distinct deliverydate  
  from orders
) d left join orders o
on o.deliverydate = d.deliverydate and not exists (
  select 1 from orders
  where memberid = o.memberid and deliverydate < o.deliverydate
)  
group by d.deliverydate