Question

我正在尝试获取订单总量，以及 30 天内的唯一客户数量。

我有这张桌子：

order_id、order_date、order_country、customer_id。

(1, '2021/03/04', 'SWE', 1),
(2, '2021/03/01', 'SWE', 1),
(3, '2021/03/01', 'DK', 3),
(4, '2021/03/01', 'DK', 3),
(5, '2021/03/03', 'NOR', 2),
(6, '2021/02/27', 'DK', 3),
(7, '2020/12/30', 'Ger', 4);

我尝试过这样的事情：

SELECT order_date
     , COUNT(order_id) as orderAmount
     , COUNT(distinct customer_id) as customerAmount 
  FROM orders
 WHERE order_date BETWEEN NOW() - INTERVAL 30 DAY AND NOW() 
 GROUP 
    BY order_date

这将获得每个日期的订单总数，这正是我想要的。但它只计算每个日期而不是每月的唯一客户，因此同一客户出现多次。我已经尝试了这个 sql 脚本的小变化，但不能让它工作。

Answer 1

请检查这是否是您想要的答案。

架构（MySQL v5.5）

create table orders (order_id int, order_date date, order_country varchar(10), customer_id int);
insert into orders values(1, '2021/03/04', 'SWE', 1);
insert into orders values(2, '2021/03/01', 'SWE', 1);
insert into orders values(3, '2021/03/01', 'DK', 3);
insert into orders values(4, '2021/03/01', 'DK', 3);
insert into orders values(5, '2021/03/03', 'NOR', 2);
insert into orders values(6, '2021/02/27', 'DK', 3);
insert into orders values(7, '2020/12/30', 'Ger', 4);

查询 #1

SELECT 
      o.order_date,COUNT(order_id) as orderAmount
     , coalesce(max(Customer_Count),0) as customerAmount 
  FROM orders o left join (select order_date,count(distinct customer_id)Customer_Count from orders o where 
 not exists (select customer_id from orders oa where o.customer_id=oa.customer_id and oa.order_date<o.order_date
            and oa.order_date > now() - INTERVAL 30 DAY  )
 group by order_date) oc on o.order_date=oc.order_date
 WHERE o.order_date BETWEEN NOW() - INTERVAL 30 DAY AND NOW() 
 group by o.order_date
 order by o.order_date;

<头>

订单日期	订单金额	客户金额
2021-02-27	1	1
2021-03-01	3	1
2021-03-03	1	1
2021-03-04	1	0

View on DB Fiddle

Answer 2

只计算客户第一次出现的时间：

SELECT order_date, COUNT(*) as num_orders,
       COUNT(DISTINCT customer_id) as distinct_customers_on_day,
       SUM(SUM(seqnum = 1)) OVER (ORDER BY order_date) as running_distinct_customers
FROM (SELECT o.*, 
             ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_date) as seqnum
      FROM orders o
      WHERE o.order_date BETWEEN NOW() - INTERVAL 30 DAY AND NOW() 
     ) o
GROUP BY order_date;

Here 是一个 db<>fiddle。

尝试计算订单总数和每月唯一客户

2 个答案: