获取一些 DISTINCT 记录的运行总计答案

【问题标题】：Getting a running total of some DISTINCT records获取一些 DISTINCT 记录的运行总计
【发布时间】：2014-01-16 15:25:24
【问题描述】：

我有一个表来跟踪我们的用户登录。我们有一个用于管理员等的仪表板，显示运行总数的图表。

我使用以下查询 (found here)：

SELECT t.date, @running_total := @running_total + t.count AS count
FROM (
    SELECT COUNT(*) AS count, DATE(datetime) AS date
    FROM user_login
    WHERE datetime >= '2013-05-01'
    GROUP BY date
) t
JOIN (
    SELECT @running_total := t2.starting_total
    FROM (
        SELECT COUNT(*) as starting_total
        FROM user_login
        WHERE datetime < '2013-05-01'
    ) t2
) initialize;

user_login 有一个 datetime 列和一个 user_id。但是，现在我还被要求显示唯一用户登录的运行总数（例如：用户 1 一天登录两次，用户 2 登录一次，即 3 次登录，但 2 次“唯一”）。我尝试这样做：

SELECT t.date, @running_total := @running_total + t.count AS count
FROM (
    SELECT COUNT(DISTINCT user_id) AS count, DATE(datetime) AS date
    FROM user_login
    WHERE datetime >= '2013-05-01'
    GROUP BY date
) t
JOIN (
    SELECT @running_total := t2.starting_total
    FROM (
        SELECT COUNT(DISTINCT user_id) as starting_total
        FROM user_login
        WHERE datetime < '2013-05-01'
    ) t2
) initialize;

但它给了我不正确的结果，我假设是因为在SELECT COUNT(DISTINCT user_id) AS count, DATE(datetime) AS date 中选择计数和日期会将其丢弃。

【问题讨论】：

标签： mysql

【解决方案1】：

您不需要如此复杂的联接来获取每个用户每天的登录次数，使用 count(*) 和 GROUP BY 将很快得到结果。

SELECT u.userid, count(*) logins_for_the_day, date(u.datetime) the_day
 FROM user_login u 
 WHERE datetime >= '2013-05-01'
 GROUP BY date(u.datetime)
 ORDER BY u.user_id, datetime

【讨论】：

【解决方案2】：

如果你想要一个累计，你为什么不这样做呢？

SELEcT date, cnt
FROM (SELECT DATE(datetime) AS date, @running_total := @running_total + count(*) as cnt
      FROM user_login cross join
           (select @running_total := 0) const
      GROUP BY date
     ) ul
WHERE datetime >= '2013-05-01';

您现在可以针对每天的唯一登录次数进行修改：

SELEcT date, cnt, cntu
FROM (SELECT DATE(datetime) AS date, @tot := @tot + count(*) as cnt,
             @totu := @totu + count(distinct user_id) as cntu
      FROM user_login cross join
           (select @tot := 0, @totu := 0) const
      GROUP BY date
     ) ul
WHERE datetime >= '2013-05-01';

注意：这给出了每天唯一登录的总和（这就是您的问题似乎在问的问题）。我怀疑您实际上想要总体上唯一登录的数量（不仅仅是一天之内）。为此，您需要一个额外的子查询来计算唯一性。这是通过查看某人登录的第一天天得出的：

SELEcT ul.date, ul.cnt, uld.cntu
FROM (SELECT DATE(datetime) AS date, @tot := @tot + count(*) as cnt
      FROM user_login cross join
           (select @tot := 0) const
      GROUP BY date
     ) ul left outer join
     (SELECT DATE(first_datetime) as date, @totu := @totu + count(*) as cntu
      FROM (select ul.user_id, min(datetime) as first_datetime
            from user_login
            group by ul.user_id
           ) ul cross join
           (select @totu := 0) const
     GROUP BY DATE(first_datetime)
    ) uld
    on uld.date = ul.date
WHERE datetime >= '2013-05-01';

【讨论】：