連續登陸天數+最大登陸天數

轉載：https://www.cnblogs.com/jiaxinwei/p/13936273.html

https://blog.csdn.net/godlovedaniel/article/details/106463180?utm_medium=distribute.pc_relevant.none-task-blog-2%7Edefault%7ECTRLIST%7Edefault-14.no_search_link&depth_1-utm_source=distribute.pc_relevant.none-task-blog-2%7Edefault%7ECTRLIST%7Edefault-14.no_search_link

 

互相是好友：https://www.cnblogs.com/alpha-cat/p/13082553.html

SQL計算最長登錄天數

計算最長登陸天數主要用兩個函數：1.窗口函數row_number()over() 2.date_sub()

1.使用row_number()窗口函數

select 
    UID,
    loadtime,
    row_number()over(partition by UID order by loadtime) sort
from 
    user_login

 

 

2.使用date_sub()函數

--取每個用戶的連續登陸天數
 1 select 
    UID,
    date_sub(loadtime,sort) as date_group, --拿登陸時間和排序相減，如果差值是定值，可以分成一組，說明用戶是連續登陸的
    min(loadtime) as start, 
    max(loadtime) as end ,
    count(1) as continue_days   --連續登陸天數
from(
    select 
        UID,
        loadtime,
        row_number()over(partition by UID order by loadtime) sort
    from 
        user_login
) a
group by 
    UID,
    date_sub(loadtime,sort)

 

3.以UID分組，取max(continue_days)

--- 用戶最大登陸天數
SELECT
    UID,
    max(continue_days) as maxday
FROM(
    select 
        UID,
        date_sub(loadtime,sort) as date_group,
        min(loadtime) as start, 
        max(loadtime) as end ,
        count(1) as continue_days
    from(
        select 
            UID,
            loadtime,
            row_number()over(partition by UID order by loadtime) sort
        from 
            user_login
        ) a
    group by 
        UID,
        date_sub(loadtime,sort)
)t1
group by UID