Subtracting Time

[EricDraven]

Does anybody know the easiest way to subtract a length of time from a give time to leave the new time?

For example :-

17:56 - 2.5 hours = 15:26

It sounds simple enought to do but I cant find anybody who can tell me a quick and simple way to accomplish this. Also when the time entered is 00:30 (for example), the system needs to know that the returned time is for the previous day. Any ideas? Anybody?

 

[Guest_imported]

It would be easy to convert the hour to minutes (from midnight) and do a simple substraction:

17:56 = 17*60+56 = 1076 minutes
2.5 hours = 2.5*60 = 150 minutes
17:56 - 2.5 = 1076 - 150 = 926 minutes

Now convert 926 minutes to hh:mm :
hh:=926 div 60; //=15
mm:=926 mod 60; //=26

If the difference is negative, then the new time is one day earlier (actually is (difference div 1440) + 1 days earlier).

Hope it's what you're looking for...

 

[svanels]

Check out the TDate and TDateTime types in the help file S. van Els
SAvanEls@cq-link.sr

 

[EricDraven]

Have checked out the TDate and TDateTime routines in the help files. They offer absolutley no help whatsoever this time but thanks anyway.

 

[Nordlund]

It depends how you want to define hours and minutes.
17:56 has a base of 60 minutes,
but the second time definition (2.5) has a base of 100.


This tiny example works if you are using correct time definitions:

procedure TForm1.Button1Click(Sender: TObject);
var
t1, t2: TTime;
begin
t1 := StrToTime('17:56');
t2 := StrToTime('2:30');
showmessage(TimeToStr(t2-t1));
end;

Andreas Nordlund
Software developer

 

[CharlesWright]

My suggestion would be to convert the minutes into a standard Delphi Date/Time and then just subtract.

To convert the minutes to a DateTime just divide the number of minutes by (60*24).

i.e.
function subtractMinutesFromDateTime(Time1 : TDateTime;Minutes : extended) : TDateTime;
var
TimeMinutes : TDateTime;
const
MinutesPerDay = 60 * 24;
begin
TimeMinutes := Minutes / MinutesPerDay;
result := time1 - TimeMinutes;
end;