Question

我需要计算一个月内订阅的人数，例如一月，当他们离开时：它看起来像：所以看起来像：

e.g。看着1月份有50位订阅者

MONTH(t2.leavedate) || COUNT(t1.subscribed1) 
January             || 10
February            || 15
March               || 3

因此，1月份队列中有50人（即1月份订阅了50人），剩下28人 - 当28人离开时，我希望我的查询输出。

这是我所拥有的表格（tableX），它显示了每个事件和user_id：

我有这个包含事件的表

id||user_id||event_name||date of event
4 ||   10  || subscribe|| 2016-01-25
5 ||   11  || subscribe|| 2016-01-30
6 ||   12  || subscribe|| 2016-01-30
7 ||   13  || leave    || 2016-02-02
8 ||   14  || subscribe|| 2016-02-03

如果我只过滤一个用户，user_id = 10（user10） - 其中user10是重新加入的示例（多个＆＃39;订阅＆＃39;事件，多个＆＃39;离开＆＃39;，表格想：

id ||user_id||event_name||date of event
 4 ||   10  || subscribe|| 2016-01-25
20 ||   10  || leave    || 2016-01-30
32 ||   10  || subscribe|| 2016-01-30
45 ||   10  || leave    || 2016-02-02
60 ||   10  || subscribe|| 2016-02-03
70 ||   10  || leave    || 2016-03-10

这是我必须计算一个月内订阅者数量的查询：

SELECT MONTHNAME(t1.joindate), COUNT(t1.subscribed1)
FROM tableX
JOIN
(SELECT tableX.user_id AS subscribed1, tableX.date_of_event AS joindate
    FROM 
        tableX
    WHERE tableX.event_name = "subscribed"
    GROUP BY tableX.id
    )t1
    ON t1.subscribed1 = tableX.user_id
WHERE event_name = 'subscribed'
GROUP BY MONTHNAME(t1.joindate)
ORDER BY t1.joindate ASC;

我遇到的问题是，如果我正在查看例如1月 - 重新加入，例如user10，在计算订阅者数量时会显示为2（COUNT（tableX.ids）） - 但是当我COUNT（t1.subscribed1）在1月份离开两次时会出现3次，而在3月份则会出现3次（总共一次，它将显示user10，作为1月份队列的一部分，离开三次，再次，user10将显示在我的订阅者计数2月队列 - 但显示用户离开3次。当我想要的2月队列，将将user10计为2月一次，并在3月离开一次

我想按顺序计算这些事件，例如2月的下一个休假日期是3月。我尝试过使用MAX-MIN和t2.leavedate＆gt; = t1.joindate等函数

这是我迄今为止所尝试过的，但它似乎并没有反映出正确的数字：

SELECT MONTHNAME(t2.leavedate), COUNT(t2.cancelled2)

    FROM tableX

JOIN 

(SELECT MONTHNAME(tableX.date_of_event), tableX.user_id AS subscribed1, tableX.date_of_event AS joindate
    FROM 
        user_account_events
    WHERE tableX.event_name = "subscribed"

    GROUP BY tableX.id
    )t1

    ON t1.subscribed1 = tableX.user_id

JOIN 

    (SELECT tableX.user_id AS cancelled2, tableX.date_of_event AS leavedate
        FROM tableX
    WHERE
        tableX.event_name = "leave"
    GROUP BY tableX.user_id
    )t2
    ON t2.cancelled2 = t1.subscribed1


WHERE tableX.date_of_event = 'leave'
AND t1.joindate BETWEEN '2016-01-01 00:00:00' AND '2016-02-01 00:00:00'
*AND t2.leavedate >= t1.joindate*

GROUP BY MONTH(t2.leavedate)
ORDER BY t2.leavedate;

1月份队列的预期输出即：（t1.joindate BETWEEN＆＃39; 2016-01-01 00：00：00＆＃39; AND＆＃39; 2016-02-01 00：00：00＆＃39; ）

MONTH(t2.leavedate) || COUNT(t1.subscribed1) 
January             || 10
February            || 15

所以，最终，我想创建一个如下所示的表：

                    Nov || Dec || Jan || Feb ||
No. of Subscribers|| 10 || 15  || 50  || 25  ||

Leave Month
              Nov || 1  ||     ||
              Dec || 3  || 2   ||
              Jan || 1  || 5   || 10 || 
              Feb || 2  || 3   || 15 || 6 ||
               ____________________________
 Total 'leaves'   || 7 || 10  || 25  || 6 ||

即。 11月份订阅的10人中有7人离开了。（11月份队列中只有3名活跃用户）

任何帮助将不胜感激！

Answer 1

此查询可让您知道某个日期是否存在“user_id”

SELECT user_id, SUM(IF(event_name="suscribe", 1, -1)) AS is_suscribed FROM `tableX` WHERE date_of_event < '2016-01-25' GROUP BY user_id

如果你想让嫌疑人在一个月内计算：

SELECT COUNT(*) FROM (SELECT user_id, SUM(IF(event_name="suscribe", 1, -1)) AS is_suscribed FROM `tableX` WHERE MONTH(date_of_event)=1 AND YEAR(date_of_event)=2016 GROUP BY user_id) AS real_suscription WHERE is_suscribed > 0

或者，如果您希望一个月内取消用户计算：

SELECT COUNT(*) FROM (SELECT user_id, SUM(IF(event_name="suscribe", 1, -1)) AS is_suscribed FROM `tableX` WHERE MONTH(date_of_event)=1 AND YEAR(date_of_event)=2016 GROUP BY user_id) AS real_suscription WHERE is_suscribed < 0

Answer 2

我认为这可能对你有好处

SELECT 
  DATE_FORMAT(next_leave_date, '%Y-%m'), 
  count(user_id) 
FROM (
  SELECT 
    tableX.user_id AS user_id, 
    MAX(tableX.date_of_event) AS next_leave_date
  FROM 
    tableX 
    JOIN ( -- get target user lists
      SELECT 
        DISTINCT user_id
      FROM 
        tableX
      WHERE date_of_event >= '2015-01-01'
        and date_of_event <  '2015-02-01'
    ) AS target_users ON tableX.user_id = target_users.user_id
    LEFT JOIN (
      SELECT 
        user_id, min(date_of_event)
      FROM
        tableX
      WHERE date_of_event >= '2015-02-01'
        AND event_name = 'subscribed'
        AND user_id IN (
          SELECT 
            DISTINCT user_id
          FROM 
            tableX
          WHERE date_of_event >= '2015-01-01'
            and date_of_event <  '2015-02-01'
        )
      GROUP BY user_id
    ) AS taget_users_with_next_sub_date ON tableX.user_id = taget_users_with_next_sub_date.user_id
  WHERE event_name = 'leave'
  GROUP BY tableX.user_id
) AS users_and_their_date_of_leave
GROUP BY DATE_FORMAT(next_leave_date, '%Y-%m');

这是一个相对复杂的SQL查询。我放在这里只能帮助您分析一个月内订阅的用户。因此，如果您想进行连贯分析，则需要多次运行，每次运行不同的月份。

我建议你在商店程序中执行此操作并将月份作为参数，这可以为您节省一些时间。

如何获取下一个日期/如何获得每次入住的时长

2 个答案: