在一个国家度过的时间的总和

Question

请考虑下表中的呼叫中心座席状态。 我需要的是计算在＆＃34; 休息＆＃34;＆＃34; 休息＆＃34;中花费的总和时间布莱恩一整天。

这是我尝试执行的内容，但它会返回一些不准确的值：

    select sum (CASE   
    WHEN State = 'Not Working' and Reason = 'Break'
    THEN Datediff(SECOND, [Time_Stamp], CURRENT_TIMESTAMP)
    else '' END) as Break_Overall
    from MyTable
    where Agent = 'Bryan'

Answer 1

使用lead()：

select agent,
       sum(datediff(second, timestamp, next_timestamp) 
from (select t.*,
             lead(timestamp) over (partition by agent order by time_stamp) as next_timestamp
      from mytable t
     ) t
where state = 'Not Working' and reason = 'Break'
group by agent;

如果代理当前可以休息，您可能需要默认值：

select agent,
       sum(datediff(second, timestamp, next_timestamp) 
from (select t.*,
             lead(timestamp, 1, current_timestamp) over (partition by agent 
                                                         order by time_stamp) as next_timestamp
      from mytable t
     ) t
where state = 'Not Working' and reason = 'Break'
group by agent;

我对这种逻辑感到有点不舒服，因为current_timestamp有一个日期组件，但你的时间不是。

编辑：

在SQL Server 2008中，您可以执行以下操作：

select agent,
       sum(datediff(second, timestamp, coalesce(next_timestamp, current_timestamp)) 
from (select t.*, t2.timestamp as next_timestamp
      from mytable t outer apply
           (select top 1 t2.*
            from mytable t2
            where t2.agent = t.agent and t2.time_stamp > t.time_stamp
            order by t.time_stamp
           ) t2
     ) t
where state = 'Not Working' and reason = 'Break'
group by agent;

Answer 2

实际上，您可以获得记录的Time_Stamp和CURRENT_TIMESTAMP之间的区别。这可能不正确 - 你可能希望得到记录的Time_Stamp和 next Time_Stamp之间相同的＆＃34; Agent＆＃34;。

（注意＆＃34;如果您有多个具有相同名称的代理，代理＆＃34;也会出现问题;您可能希望将代理存储在不同的表中并使用唯一标识符作为外键。）< / p>

所以，对于布莱恩来说，你得到了 the sum of both the "total time" for the 8:30:21 record AND the 11:34:58 record，这是对的 - 除了您正在计算＆＃34;总时间＆＃34;不正确，所以相反，你得到the sum of the time since 8:30:21 and 11:34:58。