时间戳

首页 > 解决方案 > 如何计算日期列的移动平均 DAYS

mysql - 如何计算日期列的移动平均 DAYS

问题描述

我有这些列和数据示例数据：

CREATE TABLE Orders(customerid int, orderdate datetime, orderqty int);
INSERT into Orders(customerid, orderdate, orderqty) VALUES
    (1,'2020-11-25',100),(1,'2020-11-27',160),(2,'2020-12-05',3490),
    (1,'2020-11-29',293),(2,'2020-12-07',293),(1,'2020-12-01',382);

sqlfiddle：http ://sqlfiddle.com/#!9/d90aaf/1/0

从这些数据中，我想找出每个客户在最后 3 行中的日期差异。一个示例输出是：

customerid, last 3 orders average days between orders,sum of orderqty for last 3 order
1, 2,835
2, 2,3783

我已经尝试过 datediff() 但我无法通过该函数实现此结果。

标签： mysql

解决方案

Can't figure out better trick for latest 3 orders. This works, but probably with not good performance with large tables:

SELECT o3.customerid,DATEDIFF(Latest1,IFNULL(Latest3,Latest2))/IF(Latest3 IS NULL,1,2) AS avgDiff,SUM(oQ.orderqty) AS qty3orders FROM
(SELECT o2.*,MAX(ooo.orderdate) AS Latest3 FROM 
(SELECT o1.*,MAX(oo.orderdate) AS Latest2 FROM 
(SELECT customerid,MAX(orderdate) AS Latest1 FROM Orders GROUP BY customerid) o1 
JOIN Orders oo ON o1.customerid=oo.customerid AND oo.orderdate<o1.Latest1 GROUP BY o1.customerid) o2
LEFT JOIN Orders ooo ON o2.customerid=ooo.customerid AND ooo.orderdate<o2.Latest2 GROUP BY o2.customerid) o3
JOIN orders oQ ON o3.customerid=oQ.customerid AND oQ.orderdate>=COALESCE(o3.Latest3,o3.Latest2,o3.Latest1) GROUP BY o3.customerid

推荐阅读