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

selectuser_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 continuousfrom (selectuser_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_2from (select -- 查询出绝对连续的天数user_id,flag_1,count(1) daysfrom (selectuser_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_1from 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_daysfrom (selectuser_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_idfrom (selectuser_id,login_date,datediff(login_date,prev_day) diff -- 计算与前一天的差值from(selectuser_id,login_date,lag(login_date,1,"1970-01-01") over(partition by user_id order by login_date) prev_dayfrom login) t1 ) t2 ) t3group by user_id,group_id ) t4
group by user_id

结果：

user_id	max_continuous
1001	7
1002	9

mysql/hive求最大连续天数(间隔1天也算连续)相关推荐

mysql 连续天数_mysql计算连续天数，mysql连续登录天数，连续天数统计
mysql计算连续天数,mysql连续登录天数,连续天数统计 >>>>>>>>>>>>>>>>>& ...
hive连续天数统计思路
测试数据: T1 Uid dt login_status(1登录成功,0异常) 1 2019-07-11 1 1 2019-07-12 1 1 2019-07-13 1 1 2019-07-14 1 ...
Hive 统计连续天数
第一步:创建表 -- 创建表 create table if not exists continue_days( uid int comment '员工id', tdate string commen ...
oracle求指定时间连续天数
需求:得到指定时间开始往前数连续的天数,如id为a的数据对应时间:20200115,20200114,20200112,20200101,给定查询时间为20200115,则连续天数为2,一中断就终止 ...
Ubuntu18.04 MySQL+hive安装
Ubuntu18.04 MySQL+hive安装参考文档.博客: https://blog.csdn.net/chengyuqiang/article/details/76757602 https: ...
C语言实现“输入年月求对应月份天数”
C语言实现"输入年月求对应月份天数" 一.运行截图二.源代码一.运行截图二.源代码 #include<stdio.h> main() {int x, y, a, ...
如何计算CDS view里两个时间戳之间的天数间隔
ABAP透明表里的时间戳,数据类型为dec: 有个需求:计算这两个时间戳之间的天数间隔,丢弃时间戳年-月-日8位后面的小时:分钟:秒. 举个例子:如果时间戳是20180918173132,丢弃1731 ...
hadooppythonsql_半小时搞定Hadoop+Mysql+Hive+Python
1. 说明搭建过Hadoop集群的小伙伴一定知道,如果不用docker,半小时配好Hadoop+Mysql+Hive(后简称Hive)肯定是胡吹,有了Docker镜像,没有说明文档,配好了也不一定会 ...
hive 把mysql语句执行_R分别连接mysql hive执行操作
大半年忘记更新工作记录了,惭愧,慢慢补上. mysql/hive 是两类常用的不同类型的数据库,在数据操作上存在一定的区别. 1-MYSQL的连接. library(DBI) library(RMy ...

mysql/hive求最大连续天数(间隔1天也算连续)

mysql/hive求最大连续天数(间隔1天也算连续)相关推荐

最新文章

热门文章