How to calculate difference in hours (decimal) bet

问题:

I have to calculate the difference in hours (decimal type) between two dates in SQL Server 2008.

I couldn't find any useful technique to convert datetime to decimal with 'CONVERT' on MSDN.
Can anybody help me with that?

UPDATE:
To be clear, I need the fractional part as well (thus decimal type). So from 9:00 to 10:30 it should return me 1.5.

回答1:

DATEDIFF(hour, start_date, end_date) will give you the number of hour boundaries crossed between start_date and end_date.

If you need the number of fractional hours, you can use DATEDIFF at a higher resolution and divide the result:

DATEDIFF(second, start_date, end_date) / 3600.0

The documentation for DATEDIFF is available on MSDN:

http://msdn.microsoft.com/en-us/library/ms189794%28SQL.105%29.aspx

回答2:

Just subtract the two datetime values and multiply by 24:

  Select Cast((@DateTime2 - @DateTime1) as Float) * 24.0

a test script might be:

  Declare @Dt1 dateTime Set @Dt1 = '12 Jan 2009 11:34:12'
  Declare @Dt2 dateTime Set @Dt2 = getdate()

  Select Cast((@Dt2 - @Dt1) as Float) * 24.0

This works because all datetimes are stored internally as a pair of integers, the first integer is the number of days since 1 Jan 1900, and the second integer (representing the time) is the number of (1) ticks since Midnight. (For SmallDatetimes the time portion integer is the number of minutes since midnight). Any arithmetic done on the values uses the time portion as a fraction of a day. 6am = 0.25, noon = 0.5, etc... See MSDN link here for more details.

So Cast((@Dt2 - @Dt1) as Float) gives you total days between two datetimes. Multiply by 24 to convert to hours. If you need total minutes, Multiple by Minutes per day (24 * 60 = 1440) instead of 24...

NOTE 1: This is not the same as a dotNet or javaScript tick - this tick is about 3.33 milliseconds.

回答3:

DATEDIFF but note it returns an integer so if you need fractions of hours use something like this:-

CAST(DATEDIFF(ss, startDate, endDate) AS decimal(precision, scale)) / 3600

回答4:

Using Postgres I had issues with DATEDIFF, but had success with this:

  DATE_PART('day',(delivery_time)::timestamp - (placed_time)::timestamp) * 24 + 
  DATE_PART('hour',(delivery_time)::timestamp - (placed_time)::timestamp) +
  DATE_PART('minute',(delivery_time)::timestamp - (placed_time)::timestamp) / 60

which gave me an output like "14.3"

回答5:

Declare @date1 datetime
Declare @date2 datetime

Set @date1 = '11/20/2009 11:00:00 AM'
Set @date2 = '11/20/2009 12:00:00 PM'

Select Cast(DateDiff(hh, @date1, @date2) as decimal(3,2)) as HoursApart

Result = 1.00

回答6:

You are probably looking for the DATEDIFF function.

DATEDIFF ( datepart , startdate , enddate )

Where you code might look like this:

DATEDIFF ( hh , startdate , enddate )

回答7:

DATEDIFF(minute,startdate,enddate)/60.0)

Or use this for 2 decimal places:

CAST(DATEDIFF(minute,startdate,enddate)/60.0 as decimal(18,2))

回答8:

SELECT DATEDIFF(hh, firstDate, secondDate) FROM tableName WHERE ...