mysql/hive求最大连续天数(间隔1天也算连续)

数据：

User_id	login_date
1001	2021-08-02
1001	2021-08-01
1001	2021-08-03
1001	2021-08-05
1001	2021-08-06
1001	2021-08-07
1001	2021-08-10
1001	2021-08-12
1002	2021-08-01
1002	2021-08-02
1002	2021-08-03
1002	2021-08-07
1002	2021-08-09
1002	2021-08-11
1002	2021-08-13
1002	2021-08-15

建表语句：

DROp TABLE IF EXISTS `login`;
CREATE TABLE `login`  (
  `user_id` int NOT NULL,
  `login_date` date NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = gbk COLLATE = gbk_chinese_ci ROW_FORMAT = Dynamic;

INSERT INTO `login` VALUES (1001, '2021-08-02');
INSERT INTO `login` VALUES (1001, '2021-08-01');
INSERT INTO `login` VALUES (1001, '2021-08-03');
INSERT INTO `login` VALUES (1001, '2021-08-05');
INSERT INTO `login` VALUES (1001, '2021-08-06');
INSERT INTO `login` VALUES (1001, '2021-08-07');
INSERT INTO `login` VALUES (1001, '2021-08-10');
INSERT INTO `login` VALUES (1001, '2021-08-12');
INSERT INTO `login` VALUES (1002, '2021-08-01');
INSERT INTO `login` VALUES (1002, '2021-08-02');
INSERT INTO `login` VALUES (1002, '2021-08-03');
INSERT INTO `login` VALUES (1002, '2021-08-07');
INSERT INTO `login` VALUES (1002, '2021-08-09');
INSERT INTO `login` VALUES (1002, '2021-08-11');
INSERT INTO `login` VALUES (1002, '2021-08-13');
INSERT INTO `login` VALUES (1002, '2021-08-15');

方法一：等差数列法，不灵活

select
	user_id,max(continuous) max_continuous
from (
		select 
		-- sum(days)是指间隔连续的天数,而count(*)-1是间隔连续的天数中间的值,
		-- 比如1,3,5天,sum(days)=3,而count(*)-1就是2,4这两天
		user_id,sum(days)+count(*)-1 continuous
	from (
		select
			user_id,flag_1,days,
			-- 再继续对flag_1等差,这样算出来的相同的日期就是间隔了一天的
			date_sub(flag_1,interval row_number() over(partition by user_id order by flag_1) day) flag_2
		from (
			select -- 查询出绝对连续的天数
			user_id,
			flag_1,
			count(1) days
		from (
		select
			user_id,login_date,
			-- 连续的日期flag_1是相同,而隔了一天的日期flag_1是差一天的
			DATE_SUB(login_date,interval row_number() over(partition by user_id order by login_date) day) flag_1
		from login) t1 group by user_id,flag_1
	) t2) t3 group by user_id,flag_2 )t4 group by user_id

方法二：现上分组法，灵活，可指定间隔n天连续

select
	-- 按id分组，求最大的连续天数
	user_id,
	max(continuous_login_days) max_continuous_login_days
from (
	select
		-- 将组内最大日期和最小日期相减+1就是连续天数
		user_id,datediff(max(login_date),min(login_date))+1 continuous_login_days
	from (
		select
			user_id,login_date,
			-- 超过2天以上的新开一组,因为sum默认是最前行到当前行,差值不到2天的会sum(0),就等于没加
			-- 若断2天算连续,这里就改为sum(if(diff>3,1,0))
			sum(if(diff>2,1,0)) over(partition by user_id order by login_date) group_id
		from (
			select
				user_id,login_date,
				datediff(login_date,prev_day) diff -- 计算与前一天的差值
			from(
				select
					user_id,login_date,
					lag(login_date,1,"1970-01-01") over(partition by user_id order by login_date) prev_day
			from login) t1 ) t2 ) t3
	group by user_id,group_id ) t4
group by user_id

结果：