SQL: 根据条件选择Distinct users和groupby

Question

我有一个数据库table:

User_id | User Name | Join_date  | Activity_date |  
1           abc       01/01/2021    02-01-2021
2           jay       01/01/2021    03-01-2021
2           jay       01/01/2021    04-01-2021
1           abc       01/01/2021    09-01-2021
1           abc       01/01/2021    16-01-2021 
2           jay       01/01/2021    05-01-2021
3           xyz       03/03/2021    12-03-2021
3           xyz       03/03/2021    30-03-2021
2           jay       01/01/2021    26-01-2021

我想根据他们的 Activity_date 对用户进行分桶，例如每天执行 activity 的用户分为 table1，执行 activity 的用户每周（间隔 6-7 天）进入 table2，所有其他用户进入 table3.

output 是这样的：

temporary_table1：（每隔一天做 activity 的人）

User_id | User Name | Join_date  | Activity_date |  
2           jay       01/01/2021    03-01-2021
2           jay       01/01/2021    04-01-2021
2           jay       01/01/2021    05-01-2021

temporary_table_2：（从第一次开始每 6-7 天做一次维权的人 Activity_date）

User_id | User Name | Join_date  | Activity_date |  
1           abc       01/01/2021    02-01-2021
1           abc       01/01/2021    09-01-2021
1           abc       01/01/2021    15-01-2021

temporary_table_3:(这里加上user_id=2是因为从这个用户上次activity做到最近activity做有超过7天差距)

User_id | User Name | Join_date  | Activity_date |  
3           xyz       03/03/2021    12-03-2021
3           xyz       03/03/2021    30-03-2021
2           jay       01/01/2021    26-01-2021

如何在 SQL (Redshift) 中实现这一点？

Answer 1

你应该可以做这样的事情，然后你可以在 date_diff:

上进行分组

SELECT
user_id,
user_name,
join_date
activity_date,

#window function
LAG(activity_date) OVER (PARTITION BY user_id ORDER BY activity_date ASC) as day_before,

#difference in days between activity_date and day_before
DATEDIFF(
    day, 
    LAG(activity_date) OVER (PARTITION BY user_id ORDER BY activity_date ASC),
    activity_date
    ) as date_diff
FROM your_dataset

SQL: 根据条件选择Distinct users和groupby

SQL: Selecting Distinct users and groupby based on condition

sql

group-by

amazon-redshift