SQL 中的队列分析

Question

希望对用户群进行一些同类群组分析。我们有 2 个表“用户”和“会话”，其中用户和会话都有一个“created_at”字段。我正在寻找生成一个 7 x 7 数字表（带有一些空白）的查询，该表向我显示：在特定日期创建的同时创建会话的用户计数 y = (0..6几天前），表示他那天回来了。

created_at  d2  d3  d4
today       *   *   *
today-1     49  *   *
today-2     45  30  *
today-3     47  48  18
...

在这种情况下，今天 3 日创建的 47 个用户在今天 2 日返回。

我可以在单个 MySQL 查询中执行此操作吗？我可以像这样单独执行查询，但是在一个查询中完成所有查询会非常好。

SELECT `users`.* FROM `users` INNER JOIN `sessions` ON `sessions`.`user_id` = `users`.`id` WHERE `users`.`os` = 'ios' AND (`sessions`.`updated_at` BETWEEN '2013-01-16 08:00:00' AND '2013-01-17 08:00:00')

Answer 1

这似乎是一个复杂的问题。不管它在你看来是否也是一个困难的问题，从一个较小的问题开始解决它绝不是一个坏主意。

例如，您可以从返回上周注册的所有用户（仅用户）的查询开始，即根据您的要求从现在六天后的那一天开始：

SELECT *
FROM users
WHERE created_at >= CURDATE() - INTERVAL 6 DAY

下一步可能是按日期对结果进行分组并计算每组中的行数：

SELECT
  created_at,
  COUNT(*) AS user_count
FROM users
WHERE created_at >= CURDATE() - INTERVAL 6 DAY
GROUP BY created_at

如果created_at 是datetime 或timestamp，则使用DATE(created_at) 作为分组标准：

SELECT
  DATE(created_at) AS created_at,
  COUNT(*) AS user_count
FROM users
WHERE created_at >= CURDATE() - INTERVAL 6 DAY
GROUP BY DATE(created_at)

但是，您似乎不希望输出中的 absolute 日期，而只希望 relative 日期，例如 today、today - 1 day 等。在这种情况下，您可以使用 DATEDIFF() 函数（它返回两个日期之间的天数）来生成从今天开始的（数字）偏移量并按这些值分组：

SELECT
  DATEDIFF(CURDATE(), created_at) AS created_at,
  COUNT(*) AS user_count
FROM users
WHERE created_at >= CURDATE() - INTERVAL 6 DAY
GROUP BY DATE(created_at)

您的created_at 列将包含“日期”，例如0、1 等等，直到6。将它们转换为today、today-1 等很简单，您将在最终查询中看到这一点。然而，到目前为止，我们已经到了需要后退一步（或者，也许是向右半步）的地步，因为我们实际上不需要计算用户数，而是他们的 返回。因此，目前需要的来自users 的实际工作数据集将是：

SELECT
  id,
  DATEDIFF(CURDATE(), created_at) AS day_offset
FROM users
WHERE created_at >= CURDATE() - INTERVAL 6 DAY

我们需要用户 ID 来将此行集加入（将派生自的行集）sessions，并且我们需要 day_offset 作为分组标准。

接下来，需要对sessions 表执行类似的转换，我不会对此进行详细介绍。只需说生成的查询与上一个查询完全相同，只有两个例外：

• id 被替换为 user_id；

• DISTINCT 应​​用于整个子集。

DISTINCT 的原因是每个用户和每天返回不超过一行：据我了解，无论用户在特定日期可能有多少会话，您都希望将它们计为 一次返回时间>。所以，这就是从sessions 得到的：

SELECT DISTINCT
  user_id,
  DATEDIFF(CURDATE(), created_at) AS day_offset
FROM sessions
WHERE created_at >= CURDATE() - INTERVAL 6 DAY

现在只剩下连接两个派生表，应用分组并使用条件聚合来获得所需的结果：

SELECT
  CONCAT('today', IFNULL(CONCAT('-', NULLIF(u.DayOffset, 0)), '')) AS created_at,
  SUM(s.DayOffset = 0) AS d0,
  SUM(s.DayOffset = 1) AS d1,
  SUM(s.DayOffset = 2) AS d2,
  SUM(s.DayOffset = 3) AS d3,
  SUM(s.DayOffset = 4) AS d4,
  SUM(s.DayOffset = 5) AS d5,
  SUM(s.DayOffset = 6) AS d6
FROM (
  SELECT
    id,
    DATEDIFF(CURDATE(), created_at) AS DayOffset
  FROM users
  WHERE created_at >= CURDATE() - INTERVAL 6 DAY
) u
LEFT JOIN (
  SELECT DISTINCT
    user_id,
    DATEDIFF(CURDATE(), created_at) AS DayOffset
  FROM sessions
  WHERE created_at >= CURDATE() - INTERVAL 6 DAY
) s
ON u.id = s.user_id
GROUP BY u.DayOffset
;

我必须承认我没有对此进行测试/调试，但是，如果需要，我很乐意使用您提供的数据样本，一旦您提供它们。 :)

Answer 2

这个答案反转了@Newy 想要的输出表，因此同类群组是行而不是列，并使用绝对日期而不是相对日期。

我正在寻找一个可以给我这样的查询：

Date        d0  d1  d2  d3  d4  d5  d6
2016-11-03  3   1   0   0   0   0   0
2016-11-04  4   2   0   1   0   0   *
2016-11-05  7   0   1   1   0   *   *
2016-11-06  7   3   1   1   *   *   *
2016-11-07  13  5   1   *   *   *   *
2016-11-08  4   0   *   *   *   *   *
2016-11-09  1   *   *   *   *   *   *

我正在查找某个日期注册的用户数量，然后有多少用户在 1 天后、2 天后返回等等。所以在 2016-11-07 13 用户注册并进行了会话，然后其中 5 个用户在 1 天后回来，然后有 1 个用户在 2 天后回来，依此类推。

我获取了@Andriy M 的大型查询的第一个子查询，并对其进行了修改，以提供用户注册的日期，而不是相对于当前日期的日期：

SELECT
    id,
    DATE(created_at) AS DayOffset
  FROM users
  WHERE created_at >= CURDATE() - INTERVAL 6 DAY

然后我修改的 LEFT JOIN 子查询如下所示：

 SELECT DISTINCT
    sessions.user_id,
    DATEDIFF(sessions.created_at, user.created_at) AS DayOffset
    FROM sessions
    LEFT JOIN users ON (users.id = sessions.user_id)
    WHERE sessions.created_at >= CURDATE() - INTERVAL 6 DAY

我希望 dayoffset 不是相对于@Andriy M 的答案中的当前日期，而是相对于用户注册的日期。因此，我确实在用户表上留下了联接，以获取用户注册的时间并对其进行了日期差异。

所以最终的查询看起来像这样：

SELECT u.DayOffset as Date,
  SUM(s.DayOffset = 0) AS d0,
  SUM(s.DayOffset = 1) AS d1,
  SUM(s.DayOffset = 2) AS d2,
  SUM(s.DayOffset = 3) AS d3,
  SUM(s.DayOffset = 4) AS d4,
  SUM(s.DayOffset = 5) AS d5,
  SUM(s.DayOffset = 6) AS d6
FROM (
 SELECT
    id,
    DATE(created_at) AS DayOffset
  FROM users
  WHERE created_at >= CURDATE() - INTERVAL 6 DAY
) as u
LEFT JOIN (
    SELECT DISTINCT
    sessions.user_id,
    DATEDIFF(sessions.created_at, user.created_at) AS DayOffset
    FROM sessions
    LEFT JOIN users ON (users.id = sessions.user_id)
    WHERE sessions.created_at >= CURDATE() - INTERVAL 6 DAY
) as s
ON s.user = u.id
GROUP BY u.DayOffset

Answer 3

月智队列示例：

首先让我们创建个人用户活动流表（MONTH WISE）：

SELECT 
    mu.created_timestamp AS cohort
    , mu.id AS user_id
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 1 AND l.user_id = mu.id) AS m1
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 2 AND l.user_id = mu.id) AS m2
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 3 AND l.user_id = mu.id) AS m3
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 4 AND l.user_id = mu.id) AS m4
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 5 AND l.user_id = mu.id) AS m5
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 6 AND l.user_id = mu.id) AS m6
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 7 AND l.user_id = mu.id) AS m7
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 8 AND l.user_id = mu.id) AS m8
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 9 AND l.user_id = mu.id) AS m9
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 10 AND l.user_id = mu.id) AS m10
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 11 AND l.user_id = mu.id) AS m11
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 12 AND l.user_id = mu.id) AS m12
FROM user mu 
WHERE mu.created_timestamp BETWEEN '2018-01-01 00:00:00' AND '2019-12-31 23:59:59'

然后在这个表之后计算用户的个人活动总和：

SELECT MONTH(c.cohort) AS cohort
       ,COUNT(c.user_id) AS signups
       ,SUM(c.m1) AS m1 
       ,SUM(c.m2) AS m2 
       ,SUM(c.m3) AS m3 
       ,SUM(c.m4) AS m4 
       ,SUM(c.m5) AS m5 
       ,SUM(c.m6) AS m6 
       ,SUM(c.m7) AS m7 
       ,SUM(c.m8) AS m8 
       ,SUM(c.m9) AS m9 
       ,SUM(c.m10) AS m10 
       ,SUM(c.m11) AS m11 
       ,SUM(c.m12) AS m12 
FROM (SELECT 
    mu.created_timestamp AS cohort
    , mu.id AS user_id
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 1 AND l.user_id = mu.id) AS m1
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 2 AND l.user_id = mu.id) AS m2
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 3 AND l.user_id = mu.id) AS m3
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 4 AND l.user_id = mu.id) AS m4
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 5 AND l.user_id = mu.id) AS m5
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 6 AND l.user_id = mu.id) AS m6
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 7 AND l.user_id = mu.id) AS m7
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 8 AND l.user_id = mu.id) AS m8
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 9 AND l.user_id = mu.id) AS m9
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 10 AND l.user_id = mu.id) AS m10
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 11 AND l.user_id = mu.id) AS m11
    ,(SELECT IF(COUNT(l.order_date) = 0 , 0, 1) FROM order l WHERE MONTH(l.order_date) = 12 AND l.user_id = mu.id) AS m12
FROM user mu 
WHERE mu.created_timestamp BETWEEN '2018-01-01 00:00:00' AND '2019-12-31 23:59:59') AS c GROUP BY MONTH(cohort)

您可以使用天来代替月份，其他明智的同期群分析主要用于月份情况

Answer 4

基于@Newy 回复的每月队列：

SELECT u.MonthOffset AS MONTH,

  SUM(s.MonthOffset = 0) AS m0,
  SUM(s.MonthOffset = 1) AS m1,
  SUM(s.MonthOffset = 2) AS m2,
  SUM(s.MonthOffset = 3) AS m3,
  SUM(s.MonthOffset = 4) AS m4,
  SUM(s.MonthOffset = 5) AS m5,
  SUM(s.MonthOffset = 6) AS m6
FROM (
 SELECT
    id,
    TIMESTAMPDIFF(month, DATE(date), CURDATE()) AS MonthOffset
  FROM users
  WHERE date >= CURDATE() - INTERVAL 6 month
) AS u
LEFT JOIN (
    SELECT DISTINCT
    user_id,
    TIMESTAMPDIFF(month, DATE(date), CURDATE()) AS MonthOffset
    FROM sessions
    WHERE sessions.date >= CURDATE() - INTERVAL 6 month
) AS s
ON s.user_id = u.id
GROUP BY u.MonthOffset;