建表、insert数据
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create table tmp_login ( user_id int(11) , login_date datetime);insert into tmp_login values(2,'2020-05-29 11:12:12');insert into tmp_login values(2,'2020-05-29 15:12:12');insert into tmp_login values(2,'2020-05-30 11:12:12');insert into tmp_login values(2,'2020-05-31 11:12:12');insert into tmp_login values(2,'2020-06-01 11:12:12');insert into tmp_login values(2,'2020-06-02 11:12:12');insert into tmp_login values(2,'2020-06-03 11:12:12');insert into tmp_login values(2,'2020-06-04 11:12:12');insert into tmp_login values(2,'2020-06-05 11:12:12');insert into tmp_login values(2,'2020-06-06 11:12:12');insert into tmp_login values(2,'2020-06-07 11:12:12');insert into tmp_login values(7,'2020-06-01 11:12:12');insert into tmp_login values(7,'2020-06-02 11:12:12');insert into tmp_login values(7,'2020-06-03 11:12:12');insert into tmp_login values(7,'2020-06-05 11:12:12');insert into tmp_login values(7,'2020-06-06 11:12:12');insert into tmp_login values(7,'2020-06-07 11:12:12');insert into tmp_login values(7,'2020-06-08 11:12:12'); |
方法一 row_number()
1.查询所有用户的每日登录记录
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select distinct user_id, date(login_date) as days from tmp_login; |
2.row_number()计算登录时间排序
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select user_id, days, row_number() over(partition by user_id order by days) as rnfrom ( select distinct user_id, date(login_date) as days from tmp_login) t1; |
3.用登录时间 - row_number()，如果得到的日期相同，则认为是连续登录日期
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select *, date_sub(days, interval rn day) as resultsfrom( select user_id, days, row_number() over(partition by user_id order by days) as rn from ( select distinct user_id, date(login_date) as days from tmp_login) t1) t2; |
4. 按user_id、results分组就可得出连续登录天数
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select user_id, count(*) as num_daysfrom ( select *, date_sub(days, interval rn day) as results from( select user_id, days, row_number() over(partition by user_id order by days) as rn from ( select distinct user_id, date(login_date) as days from tmp_login) t1 ) t2) t3group by user_id , results; |
直接用日期减去row_number()，不用date_sub的话，遇到登录日期跨月时会计算错误，
方法二lead() 或 lag()
这种情况适合的场景是，需要查找连续登录超过n天的用户，n为确定值
如果n为4，即计算连续登录超过4天的用户
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-- lead计算连续登录select distinct user_id from( select user_id, days, datediff(lead(days, 3, '1970-01-01') over(partition by user_id order by days), days)as results from ( select distinct user_id, date(login_date) as days from tmp_login) t1) t2where results = 3; |
连续登录4天，则日期差应该为3。
以上为个人经验，希望能给大家一个参考，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/BurningSilence/article/details/121558764
