Help with code

T

Tom Law

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• Aug 25, 2012

• #2

Try using a float value instead of a long.

 

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R

Renegade

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Longtime User

• Aug 25, 2012

• #3

for a 1 sec difference in time
z = 1000
What I am looking for is a formula that converts the variance into a decimal hr
eg
150 min = 2.5hrs
80 min = 1.3hrs

and so on

 

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thedesolatesoul

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Longtime User

• Aug 25, 2012

• #4

Renegade said:

I need the decimal to 1 decimal place for the difference in two times

Click to expand...

Can you elaborate on what you mean by '1 decimal place' in relation to time? As far as I know there are no decimal places in time unless you mean 6.5 hours or something like that.

 

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R

Renegade

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• Aug 25, 2012

• #5

It OK Solved it

B4X:

Label13.text = DateTime.Time(DateTime.Now)
   Label13.Enabled = False
   Dim x, y, z As Long
   x = DateTime.TimeParse(Label13.text)
   y = DateTime.TimeParse(Label12.text)
   z = x - y
   Label15.Text = Round2(z/60000/60,1)

/60000 converts it to min /60 converts to decimal hrs
round2( ,1) rounds to 1 decimal place

Thanks all anyway

 

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