下图为一张订单表(order1),现希望查找出至少连续3天下单的用户。
第一步:将订单表按UserId分组根据日期Date排序
第二步:用日期Date减去对应的排序号Num,若日期是连续的,则相减的结果Datedif相等。
第三步:按UserId,Datedif分组计数,得到各用户的连续下单天数。
第四步:筛选连续下单天数≥n的用户
# 订单表 order1
# UserId Date Orders
# step1: 将订单表按UserId分组根据日期Date排序
SELECT UserId,`Date`,orders,
row_number() over(PARTITION BY UserId ORDER BY DATE)Num
FROM order1
# step2: 用日期Date减去对应的排序号Num,若日期是连续的,则相减的结果Datedif相等。
SELECT UserId,`Date`,`Date`-Num AS DateDif
FROM
(
SELECT UserId,`Date`,orders,
row_number() over(PARTITION BY UserId ORDER BY DATE)Num
FROM order1
)a
# step3:按UserId,Datedif分组计数,得到各用户的连续下单天数。
SELECT UserId,COUNT(1) AS Consecutive
FROM (
SELECT UserId,`Date`,`Date`-Num AS DateDif
FROM
(
SELECT UserId,`Date`,orders,
row_number() over(PARTITION BY UserId ORDER BY DATE)Num
FROM order1
)a
)b
GROUP BY UserId,DateDif
# 筛选连续下单天数≥3的用户
SELECT Userid
FROM
(
SELECT UserId,COUNT(1) AS Consecutive
FROM (
SELECT UserId,`Date`,`Date`-Num AS DateDif
FROM
(
SELECT UserId,`Date`,orders,
row_number() over(PARTITION BY UserId ORDER BY DATE)Num
FROM order1
)a
)b
GROUP BY UserId,DateDif
)c
WHERE Consecutive>=3