在mysql中检索运行总记录随时间的增长 [英] Retrieve running-total record growth over time in mysql

问题描述

我有一个Drupal网站，有一个表跟踪用户。我想做的是图形成员随时间的增长。所以我想按下mysql返回这样的：

I have a Drupal site which has a table that keeps track of users. What I want to do is graph membership growth over time. So I want to massage mysql into returning something like this:

date | # of users (total who have registered up to the given date)
1/1/2014 | 0
1/2/2014 | 2
1/3/2014 | 10

其中＃个用户是已注册帐户的用户总数到指定日期（运行总数），而不是在该特定日期注册的用户数（这很容易检索）。

表中的 {users} 行有一个 uid 列，名称列和创建的（timestamp）列。

Each row of my {users} table has a uid column, a name column, and a created (timestamp) column.

因此，来自我的 {users} 表的示例记录将是：

So a sample record from my {users} table would be:

name: John Smith
uid: 526
created: 1365844220

推荐答案

我最终使用了一个包含变量的解决方案，基于Stack Overflow回答发布这里。此解决方案似乎比提供的其他答案更加灵活和有效。

I ended up using a solution that incorporates variables, based on a Stack Overflow answer posted here. This solution appears to be a bit more flexible and efficient than other answers provided.

  SELECT u.date,
  @running_total := @running_total + u.count AS count
  FROM (
    SELECT COUNT(*) AS count, DATE_FORMAT(FROM_UNIXTIME(created), '%b %d %Y') AS date
    FROM {users}
    WHERE created >= :start_time AND created <= :end_time
    GROUP BY YEAR(FROM_UNIXTIME(created)), MONTH(FROM_UNIXTIME(created)), DAY(FROM_UNIXTIME(created))
  ) u
  JOIN (
    SELECT @running_total := u2.starting_total
    FROM (
      SELECT COUNT(*) as starting_total
      FROM {users}
      WHERE created < :start_time
    ) u2
  ) initialize;

请注意，按组，日期格式和范围要求只是我特定项目的具体细节。 此解决方案的更通用形式（根据原始问题）是：

Note that the group by, date formatting, and range requirements are simply specifics of my particular project. A more generic form of this solution (as per the original question) would be:

  SELECT u.date,
  @running_total := @running_total + u.count AS count
  FROM (
    SELECT COUNT(*) AS count, DATE(FROM_UNIXTIME(created)) AS date
    FROM {users}
    GROUP BY date
  ) u
  JOIN (
    SELECT @running_total := 0
  ) initialize;

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