有日志如下，请写出代码求得所有用户和活跃用户的总数及平均年龄。（活跃用户指连续两天都有访问记录的用户）

数据准备

最后需完成的结果表

步骤1，所有用户的总数及平均年龄
(1). 将数据去重

with t1 as (
    select distinct
           user_id,
           age
    from test5
)

(2). 对用户总数以及平均年龄进行计算

t2 as (
select  count(user_id) as cnt,
        avg(age)       as avg_age
    from t1
)

步骤2 活跃用户的总数及平均年龄, 活跃用户指连续两天都有访问记录的用户【连续N天登录问题】

(1). 对表格进行全局的数据清洗

t3 as (
         select distinct dt,
                         user_id,
                         age
         from test5
     )

(2). 求连续活跃用户，先对每个用户进行标记。

标记方法：先将同一个用户分到一组，然后对dt进行排序，用dt减去序号。

注：一般求连续XXX的问题，基本都是用到row_number,rank等窗口函数增加标签列（伪列），然后用某列减去它们，如果是连续的，则减完后的结果是一致的，其他变相连续问题，也是该思考方式，主要是找到规律

t4 as (
         select dt,
                user_id,
                age,
                --同一个客户，按照不同日期排序，得到序号
                row_number() over (partition by user_id order by dt) as rn
         from t3
     ),
select * from t4

用dt减去rn得到天数

  t5 as (
         select *,
                --用日期减去序号得到临时日期
                date_sub(dt,rn) as date2
         from t4
     ),
select * from t5

(3). 同一个用户data2的结果是相同的，那就可以表明是连续登录的。且连续的次数大于等于2即为连续活跃用户。

    t6 as (--统计date2临时日期出现几次。如果2次则表示连续登陆2次
         select user_id,
                date2,
                count(1) as cnt
         from t5
         group by user_id,date2
         having count(1)>=2
     ),
select * from t6

(4). 一个用户可能有多次连续登录的情况，所以对上述结果去重

   t7 as (
         select distinct user_id,age
         from t6
     ),
select * from t7

(5). 求活跃用户总数

    t8 as (
         select '活跃用户'         as type,
                count(user_id) as cnt,
                avg(age)       as avg_age
         from t7
     )
     
select * from t8;

完善t2代码，并将它t2与t8合并，使用union all

   t2 as (
         select '所有用户' as type,
                count(user_id) as cnt,
                avg(age)       as avg_age
         from t1
     )

select * from t2 
union all
select * from t8;

完整代码

CREATE TABLE test5
(
    dt      string,
    user_id string,
    age     int
) ROW format delimited fields terminated BY ',';
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-11', 'test_1', 23);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-11', 'test_2', 19);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-11', 'test_3', 39);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-11', 'test_1', 23);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-11', 'test_3', 39);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-11', 'test_1', 23);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-12', 'test_2', 19);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-13', 'test_1', 23);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-15', 'test_2', 19);
INSERT INTO TABLE test_sql.test5
VALUES ('2019-02-16', 'test_2', 19);
--有日志如下，请写出代码求得所有用户和活跃用户的总数及平均年龄。（活跃用户指连续两天都有 访问记录的用户）
--步骤 1 所有用户的总数及平均年龄
with t1 as (
    select distinct
           user_id,
           age
    from test5
),
     t2 as (
         select '所有用户' as type,
                count(user_id) as cnt,
                avg(age)       as avg_age
         from t1
     ),
     --步骤 2 活跃用户的总数及平均年龄,活跃用户指连续两天都有 访问记录的用户）
     t3 as (
         select distinct dt,
                         user_id,
                         age
         from test5
     ),
     t4 as (
         select dt,
                user_id,
                age,
                --同一个客户，按照不同日期排序，得到序号
                row_number() over (partition by user_id order by dt) as rn
         from t3
     ),
     t5 as (
         select *,
                --用日期减去序号得到临时日期
                date_sub(dt,rn) as date2
         from t4
     ),
     t6 as (--统计date2临时日期出现几次。如果2次则表示连续登陆2次
         select user_id,
                date2,
                max(age) age,
                count(1) as cnt
         from t5
         group by user_id,date2
         having count(1)>=2
     ),
     t7 as (
         select distinct user_id,age
         from t6
     ),
     t8 as (
         select '活跃用户'         as type,
                count(user_id) as cnt,
                avg(age)       as avg_age
         from t7
     )
select * from t2 union all
select * from t8;