连续7天登录MySQL求解

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1 建表语句

SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;-- ----------------------------
-- Table structure for sqllevel7
-- ----------------------------
DROP TABLE IF EXISTS `orde`;
CREATE TABLE `orde`  (`id` int(11) NOT NULL,`date` datetime NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;SET FOREIGN_KEY_CHECKS = 1;

2 创建数据

INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-01 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-02 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-03 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-04 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-05 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-06 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-07 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-08 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (4, '2019-01-01 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (2, '2019-01-02 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (3, '2019-01-03 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (4, '2019-01-04 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (5, '2019-01-05 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (6, '2019-01-06 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (7, '2019-01-07 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (8, '2019-01-08 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (9, '2019-01-01 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (9, '2019-01-02 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (9, '2019-01-03 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (9, '2019-01-04 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (9, '2019-01-06 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (9, '2019-01-09 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (9, '2019-01-10 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (9, '2019-01-13 00:00:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (1, '2019-01-02 15:55:00');
INSERT INTO `test`.`sqllevel7`(`id`, `date`) VALUES (2, '2019-01-02 15:22:00');

第一步：用户登录日期去重

select DISTINCT date(date) as 日期,id from orde;

第二步：用row_number() over()函数计数

select *,row_number() over(PARTITION by id order by 日期) as cum from (select DISTINCT date(date) as 日期,id from orde)a;

第三步：日期减去计数值得到结果

select *,date(日期)-cum as 结果 from (select *,row_number() over(PARTITION by id order by 日期) as cum from (select DISTINCT date(date) as 日期,id from orde)a)b;

第四步：根据id和结果分组并计算总和，大于等于7的即为连续登陆7天的用户

select id,count(*) from (select *,date(日期)-cum as 结果 from (select *,row_number() over(PARTITION by id order by 日期) as cum from (select DISTINCT date(date) as 日期,id from orde)a)b)c GROUP BY id,结果 having count(*)>=7;

CREATE TABLE `orde`  (`id` int(11) NOT NULL,`date` datetime NULL DEFAULT NULL
)create table orde(id int,date date) row format
delimited fields terminated by ' ' ;

1 2019-01-01 00:00:00
1 2019-01-01 00:00:00
1 2019-01-02 00:00:00
1 2019-01-03 00:00:00
1 2019-01-04 00:00:00
1 2019-01-05 00:00:00
1 2019-01-06 00:00:00
1 2019-01-07 00:00:00
1 2019-01-08 00:00:00
4 2019-01-01 00:00:00
2 2019-01-02 00:00:00
3 2019-01-03 00:00:00
4 2019-01-04 00:00:00
5 2019-01-05 00:00:00
6 2019-01-06 00:00:00
7 2019-01-07 00:00:00
8 2019-01-08 00:00:00
9 2019-01-01 00:00:00
9 2019-01-02 00:00:00
9 2019-01-03 00:00:00
9 2019-01-04 00:00:00
9 2019-01-06 00:00:00
9 2019-01-09 00:00:00
9 2019-01-10 00:00:00
9 2019-01-13 00:00:00
1 2019-01-02 15:55:00
2 2019-01-02 15:22:00

hive sql

1 、连续7天登录的用户

数据:


1,2019-07-11,1
1,2019-07-12,1
1,2019-07-13,1
1,2019-07-14,1
1,2019-07-15,1
1,2019-07-16,1
1,2019-07-17,1
1,2019-07-18,1
2,2019-07-11,1
2,2019-07-12,1
2,2019-07-13,0
2,2019-07-14,1
2,2019-07-15,1
2,2019-07-16,0
2,2019-07-17,1
2,2019-07-18,0
3,2019-07-11,1
3,2019-07-12,1
3,2019-07-13,1
3,2019-07-14,0
3,2019-07-15,1
3,2019-07-16,1
3,2019-07-17,1
3,2019-07-18,1

表结构：

create table ulogin( uid int, dt date, status int )row format delimited fields terminated by ',';

导入数据:

load data local inpath '/home/hadoop/data/ulogin.dat' into table ulogin;

解决问题步骤

1.使用 row_number 在组内给数据编号(rownum)
2.某个值 - rownum = gid，得到结果可以作为后面分组计算的依据
3.根据求得的gid，作为分组条件，求最终结果

1）首先使用的row_number 给日期排序：

select uid, dt, row_number() over (partition by uid order by dt) gid from ulogin where status=1;

2）通过date_sub来获取统一的日期

select uid, dt, date_sub(dt, row_number() over (partition by uid order by dt)) gid from ulogin where status=1;

3)拿到这个结果就很简单了 group by count having

select uid,count(gid)as gg from
(select uid, dt, date_sub(dt, row_number() over (partition by uid order by dt)) gid from ulogin where status=1) as tmp
group by uid,gid
having count(gid)>=7;

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连续7天登录MySQL求解

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