Question

SELECT customer_email, count(*) AS Order_Count,

MAX(created_at) as Last_Order_Date, 
SUM(base_total_paid) AS Total_Lifetime_Sales, 
SUM(base_total_offline_refunded+base_total_online_refunded) AS Refund_Total, 

FROM mage_sales_order AS o

WHERE o.created_at > “2018-01-01” 

AND

value NOT IN (Select customer_email 
FROM mage_sales_order
WHERE WHERE o.created_at < “2018-10-01”)

试图删除上周已购买的订单，但是卡在了哪里，而且不确定！感谢您的任何帮助！

Answer 1

我认为您想删除订购<'2018-01-01'的电子邮件ID，因此它会在查询下方

SELECT customer_email, count(*) AS Order_Count,     
MAX(created_at) as Last_Order_Date, 
SUM(base_total_paid) AS Total_Lifetime_Sales, 
SUM(base_total_offline_refunded+base_total_online_refunded) AS Refund_Total     
FROM mage_sales_order AS o     
WHERE o.created_at > '2018-01-01' AND    
customer_email NOT IN (Select customer_email 
FROM mage_sales_order
WHERE  o.created_at < '2018-10-01'
)
group by customer_email

Answer 2

您需要一个group by。您的问题的答案是：

SELECT customer_email, count(*) AS Order_Count,
       MAX(created_at) as Last_Order_Date, 
       SUM(base_total_paid) AS Total_Lifetime_Sales, 
       SUM(base_total_offline_refunded+base_total_online_refunded) AS Refund_Total 
FROM mage_sales_order AS o
WHERE o.created_at < CURRENT_DATE - INTERVAL '1 week'
GROUP BY customer_email;

如果您要过滤尚未下订单的客户：

SELECT customer_email, count(*) AS Order_Count,
       MAX(created_at) as Last_Order_Date, 
       SUM(base_total_paid) AS Total_Lifetime_Sales, 
       SUM(base_total_offline_refunded+base_total_online_refunded) AS Refund_Total 
FROM mage_sales_order AS o
GROUP BY customer_email
HAVING MAX(o.created_at) < CURRENT_DATE - INTERVAL '1 week'

请注意，日期/时间函数因数据库而异，因此确切的语法可能因所使用的数据库而异。

带分组的SQL子查询

2 个答案: