Making a report with time differences.

[TrulyVisceral]

Cool, so, I'm making a simple report where I'm just showing the time difference between 2 times.
It's just =DateDiff("d",[Inicio de la interrupción de servicio],[Fin de la interrupción de servicio])

Where "Inicio" is the start and "Fin" is the end. It works alright, but as you can see in the formula, it's days.
But if the time difference is less than a day, it just shows a 0. I want to make those into hours, since there's still time there, not just 0, I want to make all the zeroes into hours, rather than days.



HOWEVER, I don't quite get the IIF command. I was thinking of adding:
=IIF([Diferencia]<1,(=DifFecha("hh:mm",[Inicio de la interrupción de servicio],[Fin de la interrupción de servicio]),"error")

CONTEXT: "Diferencia" is the field where the time difference is shown.

BUt, well, I get errors. I'm not getting the format... halp.

For now I just have a 2nd field where it is indeed shown in hours. Regardless.

[ranman256]

in the query, if days diff>0, show days, else show hours:


IIF(DateDiff("d",[Inicio de la interrupción de servicio],[Fin de la interrupción de servicio])>0,DateDiff("d",[Inicio de la interrupción de servicio],[Fin de la interrupción de servicio]),DateDiff("h",[Inicio de la interrupción de servicio],[Fin de la interrupción de servicio]))

[TrulyVisceral]

Not quite, but thanks anyways. I thought of formatting it like "hh:nn:ss".
Since that's how it's like in the excel sheet i got this data out of.

nvm, not quite once more.

[isladogs]

Date fields are stored as double number datatypes no matter how they are displayed e.g 42165.25
The integer part is the number of days since 30Dec 1899 and the decimal part is the fraction of a day.
So 0.25 is 06:00 or 6.00am

Use this fact to calculate your time diff in hours
Convert datetime values to doubles, and calculate the difference, multiply by 24 to get hours.

=24*(CDbl([End])-CDbl([Start]))

This will give results like 12.75.

If you want this to be shown as 12:45 do a further calculation to get the minutes e.g. 60*0.75

If you only want the whole number of hours, either use Round function with 0 d.p. Or the Fix function to get the completed hours.
For an answer of 12.75, these will give 13 rounded or 12 using Fix

[TrulyVisceral]

Alright. Got it. That should do it. Many thanks.

Might need help with graphics in a bit though. Let me see for myself first tho.

[isladogs]

You're welcome but ...graphics?
I draw the line at drawing clocks in Access

[TrulyVisceral]

I mean, I'll ask in a different thread.