subtracting times

[jm314]

When taking the difference between 2 times, how can I turn the value back into a time-formated string? for instance:

time1 = now
time2 = now
tdif = timevalue(time1) - timevalue(time2)

timevalue seems to use/return a double precision float. how do I convert the value of tdif to a time format string like 00:00:00 ?

 

[gmmastros]

Format(tdif, "hh:mm:ss")

-George

Strong and bitter words indicate a weak cause. - Fortune cookie wisdom

 

[SkipVought]

Hi,

Since you're calculating Elapsed Time rather than a Time-of-day value...

Format(tdif, "[h]:mm:ss")

That way, elapsed time values that exceed ONE DAY (24 hrs), will format as HOURS greater than 24.

Skip,

_{[red]Be advised:[/red] When you ignite a firecracker in a bowl of vanilla, chocolate & strawberry ice cream, you get...
Neopolitan Blownapart!}

 

[fumei]

TimeValue normally takes a string.

Dim t1
Dim t2
t1 = Now
MsgBox "gimme a sec"
t2 = Now
MsgBox Format(t2 - t1, "hh:mm:ss")

will give you the difference in time ("00:00:02" in my case), even though the variables contain full date values.

TimeValue returns a value between 00:00:00 and 23:59:59, so if you use TimeValue in the above:

Dim t1
Dim t2
Dim t3
t1 = Now
MsgBox "gimme a sec"
t2 = Now
t3 = t2 - t1
MsgBox TimeValue(Format(t3, "hh:mm:ss"))

it will return "12:00:02" - as I have my system set up as 12 hour, rather than 24 hour.

Gerry