窗口函数处理SQL连续几天登录问题(两种思路)

需求：有一张用户登录表，字段有id(用户id),time(用户登录的时间);计算连续登录天数为3天的用户。

创表语句

create table continous
(
    id INT ,
    time DATETIME  
);
INSERT into continous VALUES(201,'2017-1-1');
INSERT into continous VALUES(201,'2017-1-2');
INSERT into continous VALUES(201,'2017-1-5');
INSERT into continous VALUES(202,'2017-1-2');
INSERT into continous VALUES(202,'2017-1-3');
INSERT into continous VALUES(203,'2017-1-3');
INSERT into continous VALUES(201,'2017-1-4');
INSERT into continous VALUES(202,'2017-1-4');
INSERT into continous VALUES(201,'2017-1-5');
INSERT into continous VALUES(202,'2017-1-5');
INSERT into continous VALUES(201,'2017-1-6');
INSERT into continous VALUES(203,'2017-1-6');
INSERT into continous VALUES(203,'2017-1-7');

SELECT * FROM continous order BY id;

思路一：使用ROW_NUMBER()

第零步:去重

我们可以看到201号用户一天内登录了两次，但是我们还是得按照一天来算，所以需要先去重

select DISTINCT date(time) as time,id from continous;

第一步：根据用户分组，登陆时间排序(使用排名函数Mysql8.0后才有的)，结果按照登陆时间升序排列

SELECT ID,time,row_number() OVER(PARTITION BY ID order by time) as sort
FROM (select DISTINCT date(time) as time,id from continous) temp_0;

temp_0就是第零步得到的结果，上面整条sql语句的结果是

第二步：判断是否连续：利用登陆时间，排序得到的序列号，如果减去【使用date_sub()函数】后得到的结果是同一天，则说明是连续的；否则是不连续的

SELECT *,DATE_SUB(time,INTERVAL sort DAY) AS result FROM
(
	SELECT ID,time,row_number() OVER(PARTITION BY ID order by time) as sort
	FROM (select DISTINCT date(time) as time,id from continous) temp_0
) temp_1;

temp_1就是第一步得到的结果，上面整条sql语句的结果是

根据第二步的结果以id分组和result分组求和

SELECT id,COUNT(*) FROM
(
	SELECT *,DATE_SUB(time,INTERVAL sort DAY) AS result FROM
	(
		SELECT ID,time,row_number() OVER(PARTITION BY ID order by time) as sort --
		FROM (select DISTINCT date(time) as time,id from continous) temp_0
	) temp_1
)temp_2
GROUP BY id,result having count(*)>=1;

having后过滤的条件大于等于多少就是连续多少天登录了，我这里写的是大于等于1，就是连续登录一天以上的用户了

思路二：使用LEAD()

第零步：还是去重

-- 第0步，因为用户可能一天登录多次，去重，按一次算的
select DISTINCT date(time) as time,id from continous;

第一步：使用lead()窗口函数取它下面两行日期作为新的一列别名为next_2_time,如果下面两行为空就默认为1970年

-- temp_0就是上面第0步的结果
select id,time,lead(time,2,'1970-01-01') OVER (PARTITION BY ID ORDER BY time) next_2_time FROM 
	(
		select DISTINCT date(time) as time,id from continous
	) temp_0;

第二步：使用datediff函数(两个日期类型相减)用next_2_time减去time得到一个新列diff

SELECT *,datediff(next_2_time,time) diff FROM
(
	select id,time,lead(time,2,'1970-01-01') OVER (PARTITION BY ID ORDER BY time) next_2_time FROM 
	(
		select DISTINCT date(time) as time,id from continous
	) temp_0
) temp_1;

第三步：过滤出diff为2的就是连续登录三天的用户记录了

SELECT * FROM
	(
		SELECT *,datediff(next_2_time,time) diff FROM
		(
			select id,time,lead(time,2,'1970-01-01') OVER (PARTITION BY ID ORDER BY time) next_2_time FROM 
			(
				select DISTINCT date(time) as time,id from continous
			) temp_0
		) temp_1
	) temp_2
WHERE diff=2;

最后我们将id去重就是连续登录三天的用户了

看懂了点个赞不过分吧