Adding or subtracting time

[gvirden]

I need to calculate time differences in a database through code. What is the format for subtracting or adding time in VB?

[DiGiTaIErRoR]

I don't think there is an automated way to do it. If there is it'ld be nice! But you can do by hand(you have ten fingers and ten toes!) seriously though...
format the time into the three parts hours:minutes:seconds
then plus/minus the two times and then use If's.
like:
If DifferenceTime < 0 then
Do
Minutes = Minutes - 1
While DiffernceTime < 0
End If
It works....

------------------
DiGiTaIErRoR
VB, QBasic, Iptscrae, HTML
Quote: There are no stupid questions, just stupid people.

[HeSaidJoe]

I haven't used this so I can't explain but it's in MSDN: difftime function:

difftime
Finds the difference between two times.

double difftime( time_t timer1, time_t timer0 );

Routine Required Header Compatibility
difftime <time.h> ANSI, Win 95, Win NT


For additional compatibility information, see Compatibility in the Introduction.

Libraries

LIBC.LIB Single thread static library, retail version
LIBCMT.LIB Multithread static library, retail version
MSVCRT.LIB Import library for MSVCRT.DLL, retail version


Return Value

difftime returns the elapsed time in seconds, from timer0 to timer1. The value returned is a double-precision floating-point number.

Parameters

timer1

Ending time

timer0

Beginning time

Remarks

The difftime function computes the difference between the two supplied time values timer0 and timer1.

Example

/* DIFFTIME.C: This program calculates the amount of time
* needed to do a floating-point multiply 10 million times.
*/

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

void main( void )
{
time_t start, finish;
long loop;
double result, elapsed_time;

printf( "Multiplying 2 floating point numbers 10 million times...\n" );

time( &start );
for( loop = 0; loop < 10000000; loop++ )
result = 3.63 * 5.27;
time( &finish );

elapsed_time = difftime( finish, start );
printf( "\nProgram takes %6.0f seconds.\n", elapsed_time );
}


Output

Multiplying 2 floats 10 million times...

Program takes 2 seconds.


Floating-Point Support Routines | Time Management Routines

See Also time