Solved

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
11
361 Views
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
0
Comment
Question by:mattboy_slim
  • 6
  • 2
  • 2
  • +1
11 Comments
 
LVL 16

Expert Comment

by:brad2575
ID: 22789669
this displays all the different functions (you can go by hours, minutes, seconds, etc)

http://www.w3schools.com/vbScript/func_datediff.asp
0
 
LVL 2

Author Comment

by:mattboy_slim
ID: 22789678
I've looked at that. I'm looking for a more specific answer.
Thanks,
matt
0
 
LVL 18

Expert Comment

by:Morcalavin
ID: 22789688
Is this what your after?

Response.Write DateDiff("h", "01/01/1970 00:00:00", "01/01/1970 01:00:00")
0
 
LVL 32

Accepted Solution

by:
Daniel Wilson earned 250 total points
ID: 22789700
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

Open in new window

0
 
LVL 2

Author Comment

by:mattboy_slim
ID: 22789721
Daniel, that is exactly what I wanted, and it works perfectly. Now I just need to trim it to 2 decimal places.

Thanks!
Matt
0
Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

 
LVL 2

Author Comment

by:mattboy_slim
ID: 22789725
Morcalavin, your answer only gave the number of hours, not hours with decimal places. Thanks for the reply though.
0
 
LVL 16

Expert Comment

by:brad2575
ID: 22789731
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
0
 
LVL 2

Author Closing Comment

by:mattboy_slim
ID: 31509364
Thanks again, your code worked flawlessly.
0
 
LVL 32

Expert Comment

by:Daniel Wilson
ID: 22789761
to trim to 2 decimal places, see the formatnumber function.
http://www.w3schools.com/vbScript/func_formatnumber.asp
0
 
LVL 2

Author Comment

by:mattboy_slim
ID: 22791020
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))

%>

Open in new window

0
 
LVL 2

Author Comment

by:mattboy_slim
ID: 22791065
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)

%>

Open in new window

0

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

I recently decide that I needed a way to make my pages scream on the net.   While searching around how I can accomplish this I stumbled across a great article that stated "minimize the server requests." I got to thinking, hey, I use more than one…
I would like to start this tip/trick by saying Thank You, to all who said that this could not be done, as it forced me to make sure that it could be accomplished. :) To start, I want to make sure everyone understands the importance of utilizing p…
I designed this idea while studying technology in the classroom.  This is a semester long project.  Students are asked to take photographs on a specific topic which they find meaningful, it can be a place or situation such as travel or homelessness.…
This is a video describing the growing solar energy use in Utah. This is a topic that greatly interests me and so I decided to produce a video about it.

919 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

15 Experts available now in Live!

Get 1:1 Help Now