提交用户连续登录天数case

Author: bebee4java

cases/用户连续登录天数.sql

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+/*
+需求:
+要求统计用户连续登录的天数
+数据如下:
+uid,login_time
+1,'2016-11-25 13:30:45'
+1,'2016-11-24 13:30:45'
+1,'2016-11-24 10:30:45'
+1,'2016-11-24 09:30:45'
+1,'2016-11-23 09:30:45'
+1,'2016-11-10 09:30:45'
+1,'2016-11-09 09:30:45'
+1,'2016-11-01 09:30:45'
+1,'2016-10-31 09:30:45'
+2,'2016-11-25 13:30:45'
+2,'2016-11-24 13:30:45'
+2,'2016-11-23 10:30:45'
+2,'2016-11-22 09:30:45'
+2,'2016-11-21 09:30:45'
+2,'2016-11-20 09:30:45'
+2,'2016-11-19 09:30:45'
+2,'2016-11-02 09:30:45'
+2,'2016-11-01 09:30:45'
+2,'2016-10-31 09:30:45'
+2,'2016-10-30 09:30:45'
+2,'2016-10-29 09:30:45'
+*/
+
+-- 1. 建表
+set hive.exec.mode.local.auto=true;   --开启Hive的本地模式
+
+drop table if exists user_login;
+create table user_login(
+    uid int,
+    login_time string);
+
+insert into user_login values
+(1,'2016-11-25 13:30:45'),
+(1,'2016-11-24 13:30:45'),
+(1,'2016-11-24 10:30:45'),
+(1,'2016-11-24 09:30:45'),
+(1,'2016-11-23 09:30:45'),
+(1,'2016-11-10 09:30:45'),
+(1,'2016-11-09 09:30:45'),
+(1,'2016-11-01 09:30:45'),
+(1,'2016-10-31 09:30:45'),
+(2,'2016-11-25 13:30:45'),
+(2,'2016-11-24 13:30:45'),
+(2,'2016-11-23 10:30:45'),
+(2,'2016-11-22 09:30:45'),
+(2,'2016-11-21 09:30:45'),
+(2,'2016-11-20 09:30:45'),
+(2,'2016-11-19 09:30:45'),
+(2,'2016-11-02 09:30:45'),
+(2,'2016-11-01 09:30:45'),
+(2,'2016-10-31 09:30:45'),
+(2,'2016-10-30 09:30:45'),
+(2,'2016-10-29 09:30:45');
+
+
+-- 2. 处理
+
+select b.uid,
+min(b.login_time) login_min, --起始登录日期
+max(b.login_time) login_max, --结束登录日期
+count(b.mark_day) login_days --连续登录天数
+from (
+  -- 增加mark day
+  select a.*, date_sub(login_time, a.rank) mark_day from (
+    -- 每个用户按登录时间进行排序
+    select *,
+    rank() over(partition by uid order by login_time asc) as rank
+    from user_login
+  ) a
+) b group by b.uid, b.mark_day;
+
+
+/*
+结果:
+1	2016-10-31 09:30:45	2016-11-01 09:30:45	2
+1	2016-11-09 09:30:45	2016-11-10 09:30:45	2
+1	2016-11-24 13:30:45	2016-11-25 13:30:45	2
+1	2016-11-24 10:30:45	2016-11-24 10:30:45	1
+1	2016-11-23 09:30:45	2016-11-24 09:30:45	2
+2	2016-10-29 09:30:45	2016-11-02 09:30:45	5
+2	2016-11-19 09:30:45	2016-11-25 13:30:45	7
+*/

cases/统计某个时间段内连续登录天数>=n天的用户.sql

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+/*
+需求:
+要求统计某个时间段内连续登录天数>=5天的用户
+数据如下:
+uid,login_time
+1,'2016-11-25 13:30:45'
+1,'2016-11-24 13:30:45'
+1,'2016-11-24 10:30:45'
+1,'2016-11-24 09:30:45'
+1,'2016-11-23 09:30:45'
+1,'2016-11-10 09:30:45'
+1,'2016-11-09 09:30:45'
+1,'2016-11-01 09:30:45'
+1,'2016-10-31 09:30:45'
+2,'2016-11-25 13:30:45'
+2,'2016-11-24 13:30:45'
+2,'2016-11-23 10:30:45'
+2,'2016-11-22 09:30:45'
+2,'2016-11-21 09:30:45'
+2,'2016-11-20 09:30:45'
+2,'2016-11-19 09:30:45'
+2,'2016-11-02 09:30:45'
+2,'2016-11-01 09:30:45'
+2,'2016-10-31 09:30:45'
+2,'2016-10-30 09:30:45'
+2,'2016-10-29 09:30:45'
+*/
+
+-- 1. 建表
+set hive.exec.mode.local.auto=true;   --开启Hive的本地模式
+
+drop table if exists user_login;
+create table user_login(
+    uid int,
+    login_time string);
+
+insert into user_login values
+(1,'2016-11-25 13:30:45'),
+(1,'2016-11-24 13:30:45'),
+(1,'2016-11-24 10:30:45'),
+(1,'2016-11-24 09:30:45'),
+(1,'2016-11-23 09:30:45'),
+(1,'2016-11-10 09:30:45'),
+(1,'2016-11-09 09:30:45'),
+(1,'2016-11-01 09:30:45'),
+(1,'2016-10-31 09:30:45'),
+(2,'2016-11-25 13:30:45'),
+(2,'2016-11-24 13:30:45'),
+(2,'2016-11-23 10:30:45'),
+(2,'2016-11-22 09:30:45'),
+(2,'2016-11-21 09:30:45'),
+(2,'2016-11-20 09:30:45'),
+(2,'2016-11-19 09:30:45'),
+(2,'2016-11-02 09:30:45'),
+(2,'2016-11-01 09:30:45'),
+(2,'2016-10-31 09:30:45'),
+(2,'2016-10-30 09:30:45'),
+(2,'2016-10-29 09:30:45');
+
+
+-- 2. 处理
+-- lead函数:
+-- 第一个参数: 是指定的列（这里用登陆日期）
+-- 第二个参数: 是当前行向后几行的值，这里用的是4，也就是第五次登录的日期
+-- 第三个参数: 是如果返回的空值可以用指定值替代，这里没有使用第三个参数，返回空值
+
+select distinct uid from (
+    select uid,datediff(login_5_day, login_time) + 1 diff_day from (
+      -- 用户按login_time排序，加入第五天的登录时间
+      select *,
+      lead(login_time, 4) over(partition by uid order by login_time asc) login_5_day
+      from user_login
+    ) a
+)b where diff_day=5;
+
+/*
+结果:
+2
+*/
+-- 只有uid=2连续登录了5天以上
+
+