Question

需要找到每种电子邮件的发送日期和单击日期之间的时间差的中位数（以秒为单位）。我找到了适用于所有数据的解决方案：

SET @rowindex := -1;
SELECT g.type, g.time_diff
FROM
(SELECT @rowindex:=@rowindex + 1 AS rowindex,
TIMESTAMPDIFF(SECOND, emails_sent.date_sent, emails_clicks.date_click) AS time_diff,
emails_sent.id_type AS type
FROM emails_sent inner join emails_clicks on emails_sent.id = emails_clicks.id_email
ORDER BY time_diff) AS g
WHERE g.rowindex IN (FLOOR(@rowindex / 2) , CEIL(@rowindex / 2));

是否可以通过id_type语句添加分组？谢谢！

Answer 1

首先，您需要枚举每种类型的行。使用变量，此代码如下：

select sc.*,
       (@rn := if(@t = id_type, @rn + 1,
                  if(@t := id_type, 1, 1)
                 )
       ) as seqnum
from (select timestampdiff(second, s.date_sent, c.date_click) as time_diff,
             s.id_type,
      from emails_sent s inner join
           emails_clicks c
           on s.id = c.id_email
      order by time_diff
     ) sc cross join
     (select @t := -1, @rn := 0) as params;

然后，您需要输入每种类型的总数并计算中位数：

select sc.id_type, avg(time_diff)
from (select sc.*,
             (@rn := if(@t = id_type, @rn + 1,
                        if(@t := id_type, 1, 1)
                       )
             ) as seqnum
      from (select timestampdiff(second, s.date_sent, c.date_click) as time_diff,
                   s.id_type,
            from emails_sent s inner join
                 emails_clicks c
                 on s.id = c.id_email
            order by time_diff
           ) sc cross join
           (select @t := -1, @rn := 0) as params
     ) sc join
     (select id_type, count(*) as cnt
      from emails_sent s inner join
           emails_clicks c
           on s.id = c.id_email
      group by id_type
     ) n
where 2 * seqnum in (n.cnt, n.cnt, n.cnt + 1, n.cnt + 2)
group by sc.id_type;

如何使用group by（MySQL）查找中间值

1 个答案: