Multiplying with DateDiff result (1 Viewer)

Sketchin · Sep 18, 2012

I am using datediff to calculate how many hours are between 2 times and getting a result like 7:30.

Is there a way to use that result in a calculation to figure out how many hours occured in so many days?

This is my current expression:
TrainingManHours: (weekdays([TrainingStartDate],[TrainingEndDate]))*(DateDiff("h",[StartTime],[EndTime]))

Btw..weekdays is just a function that tells me how many weekdays are between 2 dates.

pr2-eugin · Sep 18, 2012

I thought it would return the integer.. I just used the following in the immediate window to double check it gave the right answer..

Code:

? DateDiff("h",#18/09/2012 09:14:23 AM#,Now)
 7

AccessMSSQL · Sep 18, 2012

Can you just cast both sides of that equation to integer and then you won't get the 7:30 result. What type is your weekdays function returning?

Sketchin · Sep 18, 2012

I must've confused myself because I did a test with Format, and it returned 7:30

test: Format([StartTime]-1-[EndTime],"Short Time")

Will look into this further, thanks for your time.

(Weekdays returns an integer)

Sketchin · Sep 18, 2012

Ok, my problem was if I do
(DateDiff("h",[StartTime],[EndTime]))
and my start time is 8:30 and end time is 4:00, the result returned is 8, not 7.5.

I assume this is because of my data type being integer rather than ...double?

pr2-eugin · Sep 18, 2012

Even if the variable is double you will still get that same value, because the value returned by DateDiff function is an Integer.

Sketchin · Sep 18, 2012

The solution was to do this:

test: (([StartTime]-[EndTime])*-24)*(weekdays([TrainingStartDate],[TrainingEndDate]))

Multiplying with DateDiff result (1 Viewer)

Sketchin

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pr2-eugin

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AccessMSSQL

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Sketchin

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Sketchin

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pr2-eugin

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Sketchin

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