Question

我的表有4列：branch_name，customer_id，order_value，purchase_date。

"london-branch", 422, "12","01-01-2019"
"manchester", 133, "33","01-04-2019"
"london", 422, "55","01-04-2019"
"newyork", 1223, "11","01-04-2019"

想要找出2019年花了超过2018年总花销额10％以上的顾客。

我尝试过

CREATE TABLE AS LastYr 

Select customer_id,order_value from order_table where purchase_date between '01-01-2018' AND '31-12-2018';

这将为2018年创建表格

然后

CREATE TABLE AS NewYear

Select customer_id,order_value from order_table where purchase_date between '01-01-2019' AND '31-12-2019';

然后努力拼凑桌子，找到不同之处。

Answer 1

您可以在SUM子句中使用聚合和条件HAVING：

SELECT customer_id
FROM mytable
WHERE YEAR(purchase_date) IN (2018, 2019)
GROUP BY customer_id
HAVING
    SUM(CASE WHEN YEAR(purchase_date) = 2019 THEN order_value ELSE 0 END)
    > 1.1 * SUM(CASE WHEN YEAR(purchase_date) = 2018 THEN order_value ELSE 0 END)

注意：这假设您使用相关数据类型而不是字符串存储日期。

寻找2019年花费比2018年至少多10％的客户

1 个答案: