Time difference for leave application (1 Viewer)

[gsrajan]

Please let me know what is syntax to get the total time used in hours and minutes:

1. Start time: 9.30 am
End time: 5.00 pm
I need to get used time as 7.50 ( that is 71/2 hours)

2. Start time: 9.30 am
End time: 5.30 pm
I need to get used time as 8.00

Thanks for your help.

 

[Minty]

Use the DateDiff() function - https://docs.microsoft.com/en-us/office/vba/language/reference/user-interface-help/datediff-function

MyValue: Datediff("n",[Timefield1],[TimeField2]) / 60

Should get you started

 

[gsrajan]

Thanks so much.

 

[gsrajan]

Thanks so much.

I tried this:

Start time: 10.30 A.M. End time 12.00 P.M.
I get the output 2. Not 1.5.

This is my code:

txtTotal = DateDiff("n", [StartTime], [EndTime]) / 60

Thank you.

 

[Minty]

I don't know why , in the immediate window if I type

? DateDiff("n", DATE(), Now()) / 60

I get

20.6666666666667

Did you define txtTotal as an integer?

 

[Gasman]

What type is txtTotal?

starttime = #10:30 AM#
endtime= #12:00 PM#
txtTotal = DateDiff("n", StartTime, EndTime) / 60
? txttotal
 1.5

 

[gsrajan]

Yes, I tried your above code and tried this also. Getting same result. Not sure why. Thanks for trying.

Dim myInt As Integer
myInt = DateDiff("n", StartTime, EndTime) / 60
txtTotal = myInt

 

[Gasman]

Yes, I tried your above code and tried this also. Getting same result. Not sure why. Thanks for trying.

Dim myInt As Integer
myInt = DateDiff("n", StartTime, EndTime) / 60
txtTotal = myInt

How on earth do you expect to get a value like 1.5 when you are using an integer type?

 

[gsrajan]

I just said apart from the code your provided, this is also not working. Can you please tweak it if you can? thanks

 

[Minty]

An integer data type can only store a integer value - a whole number.
Pick a data type that has some decimal portion;

Dim MyResult as Currency
MyResult = DateDiff("n", StartTime, EndTime) / 60