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how can i get the result of two time.

Posted on 2003-11-11
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Last Modified: 2010-04-06
how can i get the result of two time.
sample

7:05 AM - 8:30 AM
or
5:48 PM - 7:30 PM

thanks
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Question by:ziaguintu6
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Expert Comment

by:Dean OBrien
ID: 9721202
what language
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LVL 35

Expert Comment

by:YZlat
ID: 9721922
VBScript example:

use DateDiff function.
DateDiff(interval, number, date)
Where interval could be one of those:

h      ---> Hour
n      ---> Minute
s      ---> Second

Example:
Dim diffInMinutes, diffInSec
diffInMinutes=DateDiff("n",8:30, 7:05)
diffInSec=DateDiff("n",8:30, 7:05)




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Expert Comment

by:jimmack
ID: 9721931
Do you mean you want the difference between the two times?

In Java:

        Calendar startTime = Calendar.getInstance();
        startTime.clear();
        startTime.set(2003, 10, 11, 7, 5);
        Calendar endTime = Calendar.getInstance();
        endTime.clear();
        endTime.set(2003, 10, 11, 8, 30);
       
        long diff = endTime.getTimeInMillis() - startTime.getTimeInMillis();
       
        System.out.println("diff = " + diff);

This give 5100000 (5100 seconds = 85 minutes = 1 hour 25 minutes).

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Expert Comment

by:Zontar
ID: 9726234
JavaScript:

var today = new Date();
var year = today.getFullYear();
var month = today.getMonth();
var day = today.getDay();

var start = new Date(year, month, day, sHours, sMinutes, sSeconds).getTime();
var end = new Date(year, month, day, eHours, eMinutes, eSeconds).getTime();

//  you supply the hours, minutes, seconds for each

var diff = (end - start) * 1000; //  difference in seconds
var diffHours = Math.floor(diff / 3600);
var diffMinutes = Math.floor( (diff - (diffHours * 3600)) / 60 );
var diffSeconds = Math.floor(diff - (diffHours * 3600) - (diffMinutes * 60));

PHP:

$start = mktime($s_hours, $s_minutes, $s_seconds);
$end = mktime($e_hours, $e_minutes, $e_seconds);

# you supply the hours. minutes, seconds for each

$diff = $end - $start;

$diff_formatted = date("G:i:s", $diff);

MySQL (pre-4.1.1):

SELECT DATE_SUB("2003-11-12 17:42:35", INTERVAL "12:30:20" HOUR_SECOND) AS difference;

(version 4.1.1 and higher)

SELECT SUBTIME("17:42:35", "12:30:20") AS difference;
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Author Comment

by:ziaguintu6
ID: 9727236
i am using asp and i can't understand your answer. i assigned the two dates from two variables

thanks
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Expert Comment

by:jimmack
ID: 9727540
LOL: About the only thing no one covered ;-)

I can't help with that :-(
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Accepted Solution

by:
Zontar earned 25 total points
ID: 9727693
Well, you could just use JScript for your ASP (and the code I posted above). But since I enjoy suffering, here it is in ASP/VBScript ... ;^)

<%
  Time1 = TimeValue("3:45:30AM")
  Time2 = TimeValue("5:25:40PM")

  H1 = Hour(Time1)
  M1 = Minute(Time1)
  S1 = Second(Time1)

  H2 = Hour(Time2)
  M2 = Minute(Time2)
  S2 = Second(Time2)

  Diff = TimeSerial(H2 - H1, M2 - M1, S2 - S1)

  HD = Hour(Diff)
  MD = Minute(Diff)
  SD = Second(Diff)

  Response.Write HD & ":" & MD & ":" & SD
  ' Displays: 13:40:10
%>
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Assisted Solution

by:YZlat
YZlat earned 25 total points
ID: 9731060
<%
Dim diffInMinutes, diffInSec
diffInMinutes=DateDiff("n",8:30, 7:05)
diffInSec=DateDiff("n",8:30, 7:05)

%>
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Expert Comment

by:YZlat
ID: 9731065
The diffInMinutes=DateDiff("n",8:30, 7:05)
 will return the difference in minutes and diffInSec=DateDiff("n",8:30, 7:05)
will return the difference in seconds
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Expert Comment

by:Zontar
ID: 9732228
I didn't realise you could pass DateDiff a dummy argument like that, thanks for the tip.

I usually do all my ASP stuff in JScript (or occasionally ActivePython) and only use VBScript when someone holds a gun to my head. ;^)
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