SQL 当有多个具有最早日期的行时按最小日期分组

Question

我正在尝试提出一个查询，该查询将返回客户最早下订单的聚合数据。 当给定客户的最早购买日期有多个行/订单时，是否可以这样做？ 例如，我有这些表：

                                 customers

ID	姓名	created_at
1	山姆	2019-07-12
2	吉米	2019-01-22

                                   items

ID	姓名	价格
1	手表	200
2	腰带	75
3	钱包	150

                                  orders

ID	客户ID	item_id	created_at
1	1	1	2018-08-01
2	1	2	2018-08-11
3	2	1	2019-01-22
4	2	3	2019-01-22
5	2	2	2019-03-03

                             expected query

客户ID	姓名	first_purchase_date	n_items	总价
1	山姆	2018-08-01	1	200
2	吉米	2019-01-22	2	350

我目前设置了以下查询，但此查询按 customer_id 分组，因此商品总数和总价格不反映最早的订单。

SELECT 
    orders.customer_id, 
    customers.name AS name, 
    MIN(orders.created_at) AS first_purchase_date,
    COUNT(*) as n_items,
    SUM(items.price) as total_price
FROM orders
INNER JOIN customers
    ON orders.customer_id = customers.id
INNER JOIN items
    ON orders.item_id = items.id
GROUP BY
    customers.id

         my incorrect query

客户ID	姓名	first_purchase_date	n_items	总价
1	山姆	2018-08-01	2	275
2	吉米	2019-01-22	3	425

非常感谢任何帮助。 谢谢！

Answer 1

解释：

• cte将为您提供每个客户 ID 和名称的 first_purchase_date
• 使用该表查找与客户 ID 和 first_purchase_date 匹配的聚合

见分贝<>小提琴

WITH cte AS (
SELECT 
    orders.customer_id, 
    customers.name, 
    MIN(orders.created_at) AS first_purchase_date
FROM orders
INNER JOIN customers
    ON orders.customer_id = customers.id
INNER JOIN items
    ON orders.item_id = items.id
GROUP BY
    orders.customer_id,
    customers.name
)

SELECT
    cte.customer_id,
    cte.name,
    cte.first_purchase_date,
    COUNT(orders.item_id) AS n_items,
    SUM(items.price) AS total_price
FROM cte
INNER JOIN orders 
    ON cte.customer_id = orders.customer_id
    AND cte.first_purchase_date = orders.created_at
INNER JOIN items
    ON orders.item_id = items.id
GROUP BY 
    cte.customer_id,
    cte.name,
    cte.first_purchase_date
;

Answer 2

select   customer_id
        ,name
        ,created_at as first_purchase_date
        ,count(*)   as n_items
        ,sum(price) as total_price
from    (
         select    o.customer_id
                  ,c.name
                  ,o.created_at
                  ,i.price
                 ,rank() over(partition by o.customer_id order by o.created_at) as rn
         from     orders o join items i on i.id = o.item_id join customers c on c.id = o.customer_id
        ) t
where    rn = 1
group by customer_id, name, created_at

客户ID	姓名	first_purchase_date	n_items	总价
1	山姆	2018-08-01 00:00:00	1	200
2	吉米	2019-01-22 00:00:00	2	350

小提琴