Question

我想在00-00-00 00:00:00得到日期差异当我减去2个日期时，我得到的结果总是从1900-01-01开始。

如何才能显示剩余的日期和时间。或甚至表示（X）天（X）小时（X）分钟

SELECT TOP 
   [Open_Date]
  ,[Close_Date]   
  ,([Open_Date]-[Close_Date]) as [test]
  ,[time_spent_sum]
  ,[support_lev]
  ,[Priority]
  ,[CountCases]
  ,[Last_Update_Status]
  ,[Last_Update_Date]
  ,[Customer Group]
  ,[Assignee Group]
  ,[3rd_Party]
  ,[rootcause]
  ,[location_name]
  ,[Dataline]
  ,[InterfaceID]
  ,[DAYS OPEN]
  FROM [mdb].[dbo].[ReportsView]

我一直在捣乱一些皈依者，但没有达到预期的效果。

case
  when [Close_Date] is null then format(convert(smalldatetime, dateadd(minute,  (datediff(minute,[Open_Date],getdate())), ''), 114),'d ''days''   HH ''Hrs'' :mm ''Mins'' ') 
  when [Close_Date] is not null then format(convert(smalldatetime, dateadd(minute,  (datediff(minute,[Open_Date],[Close_Date])), 0), 114),'d ''days''   HH ''Hrs'' :mm ''Mins'' ') 
  else null 
  end as [Total_call_open_time]

DATES

Answer 1

为了更准确地表示两个时间段之间的天数/小时/分钟/秒，您需要找到两个日期之间的实际时间，因为datediff实际上告诉您跨越了多少个边界。

例如，即使两个时期差异接近48小时，以下两者都会返回值1：

select datediff(day,'2000-01-01 00:00:01','2000-01-02 23:59:59')
select datediff(day,'2000-01-01 23:59:59','2000-01-02 00:00:01')

因此，以下内容可以满足您的需求，使用div /和mod %来计算总时间跨度和剩余时间：

declare @d1 datetime = dateadd(day,-1,dateadd(hour,-2,dateadd(minute,-3,dateadd(second,-4,getdate()))))
declare @d2 datetime = getdate()

select datediff(hour,@d1,@d2)/24 as Days
        ,datediff(minute,@d1,@d2)/60%24 as Hours
        ,datediff(second,@d1,@d2)/60%60 as Minutes
        ,datediff(second,@d1,@d2)%60%60 as Seconds

        ,cast(datediff(hour,@d1,@d2)/24 as nvarchar(3)) + case when cast(datediff(day,@d1,@d2) as nvarchar(3)) = 1 then ' Day, ' else ' Days, ' end
         + cast(datediff(minute,@d1,@d2)/60%24 as nvarchar(3)) + case when cast(datediff(hour,@d1,@d2)%24 as nvarchar(3)) = 1 then ' Hour, ' else ' Hours, ' end
         + cast(datediff(second,@d1,@d2)/60%60 as nvarchar(3)) + case when cast(datediff(Minute,@d1,@d2)%60 as nvarchar(3)) = 1 then ' Minute, ' else ' Minutes, ' end
         + cast(datediff(Second,@d1,@d2)%60%60 as nvarchar(3)) + case when cast(datediff(Second,@d1,@d2)%60%60 as nvarchar(3)) = 1 then ' Second' else ' Seconds' end as TimeDifference

如何通过从另一个日期中减去一个日期来获得日期结果

1 个答案: