Question

与SQL - Select most 'active' timespan fromdb非常密切相关，但问题不同。

“我有一个交易表。在此表中，我将交易日期时间存储在UTC中。我有几个月的数据，每天约20,000笔交易。”

如何改变

  select datepart(hour, the_column) as [hour], count(*) as total 
  from t 
  group by datepart(hour, the_column) 
  order by total desc

这样我就可以选择最活跃的特定年份，月份，日期，时间，分钟和秒。

为了澄清，我不是在寻找当天哪个小时或哪个时段最活跃。相反，哪个时刻最活跃。

Answer 1

Select 
    DATEPART(year, the_column) as year
    ,DATEPART(dayofyear,the_column) as day
    ,DATEPART(hh, the_column) as hour
    ,DATEPART(mi,the_column) as minute
    ,DATEPART(ss, the_column) as second
    ,count(*) as count from t
Group By 
    DATEPART(year, the_column)
    , DATEPART(dayofyear,the_column)    
    , DATEPART(hh, the_column)
    , DATEPART(mi,the_column)
    , DATEPART(ss, the_column)
order by count desc

Answer 2

如果分钟分辨率足够：

select top 1 cast(the_column as smalldatetime) as moment, count(*) as total 
from t 
group by cast(the_column as smalldatetime)
order by total desc

SQL - 选择最活跃的＆＃39;来自db的时间

2 个答案: