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ASP Math: Get amount of time between two values (such as 10/23/2008 2:30:00 and 10/23/2008 4:30:00)?

Posted on 2008-10-23
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Last Modified: 2012-05-05
Using Classic ASP, I need to get the amount of time (in hours) between two values:
For example:
 - 10/23/2008 2:00:00 to 10/23/2008 2:20:00 = 0.33
 - 10/23/2008 7:45:00 to 10/23/2008 11:30:00 = 3.75

I've researched 'datediff' and 'dateadd' with no luck.  I cannot seem to find any examples of 'datediff' using the time, only the date, but apparently you get get number of seconds/minutes/hours.

Thanks in advance,
Matt
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Question by:mattboy_slim
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Expert Comment

by:brad2575
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this displays all the different functions (you can go by hours, minutes, seconds, etc)

http://www.w3schools.com/vbScript/func_datediff.asp
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by:mattboy_slim
Comment Utility
I've looked at that. I'm looking for a more specific answer.
Thanks,
matt
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Expert Comment

by:Morcalavin
Comment Utility
Is this what your after?

Response.Write DateDiff("h", "01/01/1970 00:00:00", "01/01/1970 01:00:00")
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Daniel Wilson earned 250 total points
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http://www.w3schools.com/vbScript/func_datediff.asp




dim iNumSecs

dim dHoursDiff
 

iNumSecs = datediff("s", "10/23/2008 2:00:00", " 10/23/2008 2:20:00")

dHoursDiff = cdbl(iNumSecs / 3600.0)
 

response.write dHoursDiff

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by:mattboy_slim
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Daniel, that is exactly what I wanted, and it works perfectly. Now I just need to trim it to 2 decimal places.

Thanks!
Matt
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Author Comment

by:mattboy_slim
Comment Utility
Morcalavin, your answer only gave the number of hours, not hours with decimal places. Thanks for the reply though.
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Expert Comment

by:brad2575
Comment Utility
if you want to convert something to a time in minutes, etc, you would take the datediff useing the seconds and then do some math on it to get minutes, hours, etc.

So you would take
 this will give you ONLY the seconds left (remainder from getting minuts)
Seconds = DateDiff("s", "01/01/1970 00:00:00", "01/01/1970 01:00:00")/60 - DateDiff("m", "01/01/1970 00:00:00", "01/01/1970 01:00:00")

 this will give you ONLY minutes left (remainder from getting hours)
Minutes = DateDiff("h", "01/01/1970 00:00:00", "01/01/1970 01:00:00")/60 - DateDiff("m", "01/01/1970 00:00:00", "01/01/1970 01:00:00")

 this will give you ONLy hours left (remainder from getting days
Hours = DateDiff("d", "01/01/1970 00:00:00", "01/01/1970 01:00:00")/24 - DateDiff("h", "01/01/1970 00:00:00", "01/01/1970 01:00:00")

-- so on
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Author Closing Comment

by:mattboy_slim
Comment Utility
Thanks again, your code worked flawlessly.
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Expert Comment

by:Daniel Wilson
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to trim to 2 decimal places, see the formatnumber function.
http://www.w3schools.com/vbScript/func_formatnumber.asp
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by:mattboy_slim
Comment Utility
Thanks again Daniel, it works great.  Here is the code that works for anyone searching in the future:
<%    

dim var_DateDiffSeconds, var_DateDiffHours, var_datefirst, var_datenext

var_datefirst = request.querystring("varfirst")

var_datenext = request.querystring("varnext")

 

var_DateDiffSeconds = datediff("s", var_datefirst, var_datenext)

var_DateDiffHours = cdbl(var_DateDiffSeconds / 3600.0)

 

response.write var_DateDiffHours & "<br/>"

response.write("Hours: " & FormatNumber(var_DateDiffHours, 2))

%>

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Author Comment

by:mattboy_slim
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I compressed it a little more:
<%    

dim var_DateDiffSeconds, var_DateDiffHours, var_datefirst, var_datenext

var_datefirst = request.querystring("varfirst")

var_datenext = request.querystring("varnext")

 

var_DateDiffTotal = FormatNumber(cdbl(datediff("s", var_datefirst, var_datenext) / 3600.0), 2)
 

response.write("Hours: " & var_DateDiffTotal)

%>

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