1

假设我有我的table喜欢：

uid  day_used_app   
---  -------------
1    2012-04-28      
1    2012-04-29        
1    2012-04-30        
2    2012-04-29       
2    2012-04-30 
2    2012-05-01       
2    2012-05-21        
2    2012-05-22

2012-05-03假设我想要在过去 7 天（从）中至少 2 天返回应用的唯一用户数。

例如，检索过去 7 天内至少 2 天使用过该应用程序的用户数：

select count(distinct case when num_different_days_on_app >= 2
                           then uid else null end) as users_return_2_or_more_days

    from (
         select uid,
                count(distinct day_used_app) as num_different_days_on_app
             from table
         where day_used_app between current_date() - 7 and current_date()
         group by 1

        )

这给了我：

users_return_2_or_more_days
---------------------------
            2

我的问题是：

如果我想到目前为止的每一天都这样做，我的表看起来像这样，其中第二个字段等于在第一个字段中的日期之前一周内返回 2 个或更多不同天的唯一用户数。

        date               users_return_2_or_more_days
      --------             ---------------------------
    2012-04-28                        2
    2012-04-29                        2 
    2012-04-30                        3           
    2012-05-01                        4     
    2012-05-02                        4       
    2012-05-03                        3
4

2 回答 2

1

这会有帮助吗？

WITH
-- your original input, don't use in "real" query ...
input(uid,day_used_app) AS (
          SELECT 1,DATE '2012-04-28'
UNION ALL SELECT 1,DATE '2012-04-29'
UNION ALL SELECT 1,DATE '2012-04-30'
UNION ALL SELECT 2,DATE '2012-04-29'
UNION ALL SELECT 2,DATE '2012-04-30'
UNION ALL SELECT 2,DATE '2012-05-01'
UNION ALL SELECT 2,DATE '2012-05-21'
UNION ALL SELECT 2,DATE '2012-05-22'
)
-- end of input, start "real" query here, replace ',' with 'WITH'
,
one_week_b4 AS (
  SELECT
    uid
  , day_used_app
  , day_used_app -7 AS day_used_1week_b4
  FROM input
)
SELECT
  one_week_b4.uid
, one_week_b4.day_used_app
, count(*) AS users_return_2_or_more_days
FROM one_week_b4
JOIN input
  ON input.day_used_app BETWEEN one_week_b4.day_used_1week_b4 AND one_week_b4.day_used_app
GROUP BY
  one_week_b4.uid
, one_week_b4.day_used_app
HAVING count(*) >= 2
ORDER BY 1;

输出是：

uid|day_used_app|users_return_2_or_more_days
  1|2012-04-29  |                          3
  1|2012-04-30  |                          5
  2|2012-04-29  |                          3
  2|2012-04-30  |                          5
  2|2012-05-01  |                          6
  2|2012-05-22  |                          2

这对您的需求有帮助吗？

理智的马可...

于 2017-05-26T20:33:54.537 回答
0 
SELECT DISTINCT
    t1.day_used_app,
    (
        SELECT SUM(CASE WHEN t.num_visits >= 2 THEN 1 ELSE 0 END)
        FROM
        (
            SELECT uid,
                   COUNT(DISTINCT day_used_app) AS num_visits
            FROM table
            WHERE day_used_app BETWEEN t1.day_used_app - 7 AND t1.day_used_app
            GROUP BY uid
        ) t
   ) AS users_return_2_or_more_days
FROM table t1
于 2017-05-26T05:56:48.600 回答