Question

我的数据如下：

TicketID    Status    Datetime
25          Open      2016-07-21 11:46:40
25          Open      2016-07-22 09:42:40
25          Closed    2016-07-22 09:46:40
25          Open      2016-07-23 08:42:40
25          Open      2016-07-24 08:46:40
25          Open      2016-07-24 09:26:40
25          Closed    2016-07-25 11:46:40

我想计算以下内容：

<（>（2016-07-22 09:46:40 - 2016-07-21 11:46:40）+（2016-07-25 11:46:40 - 2016-07-23 08:42:40 ））/ 2。

Answer 1

这是一个想法：对于每个“开放”获得下一个“关闭”。然后只选择“关闭”值并采用最小值打开以确定时间长度。因此，以下内容可以获得您想要的一对次数：

select close_datetime, min(datetime) as open_datetime
from (select t.*,
             (select t2.DateTime
              from t t2
              where t2.status = 'Closed' and t2.DateTime > t.DateTime
             ) as close_datetime
      from t
      where status = 'Open'
     ) t
group by close_datetime;

然后您可以获得差异的总和。这将以秒为单位获取值：

select sum(timestampdiff(second, clsoe_datetime, open_datetime) as num_seconds
from (select close_datetime, min(datetime) as open_datetime
      from (select t.*,
                   (select t2.DateTime
                    from t t2
                    where t2.status = 'Closed' and t2.DateTime > t.DateTime
                   ) as close_datetime
            from t
            where status = 'Open'
           ) t
      group by close_datetime
     ) t

Answer 2

这对我来说似乎没问题。

set @x=now(), @y=-1;

select 
    avg(timestampdiff(second,prev_stat_dt,curr_stat_dt))
from
    (select 
        TicketID,
            @y prev_stat,
            @y:=Status curr_stat,
            @x prev_stat_dt,
            @x:=DateTime curr_stat_dt
    from
        table1
    where
           Status in ('Open' , 'Closed')) t
where
    prev_stat = 'Open' and curr_stat = 'Closed';

如何计算MySQL中的解析时间？

2 个答案: