code to generate random integer variable 1

[haytatreides]

does anyone know code to generate a random integer for a variable??? i want to have an if statement execute 20% of the time, I figure that if i can get a variable to take an integer from 1-100 and if that number is 1-20 then execute. anyone know how i can do this?

 

[LPlates]

Private Sub Command1_Click()
Dim RndmInt As Integer

Randomize
RndmInt = Int(Rnd * 100)
End Sub

 

[VBAjedi]

Since Rnd() generates a number between 0 and 1, you could just use this for your "If" statement:

If Rnd() < 0.21 Then
   ' Do whatever
End If

VBAjedi

 

[haytatreides]

that did it, thanks

 

[haytatreides]

ok, i lied.

it didn't work. i tried the format

MyValue = Cint(Int((101 * Rnd()) + 1))

and it is telling me that the object method isn't supported. I am guesssing because rnd() is a .net function and access 2002 uses VB6.3 could that be it? or am i just doing it wrong?

hayt

 

[atray04]

your guessing? Because you could just do a F1 on rnd() to see if its in the vba help files and look at it there.

 

[VBAjedi]

hayt,

Your code works fine in Excel VBA. . . but I'm not seeing why you want to take that approach. Unless you need a random number between 1 and 100 for something else?

The way I used Rnd() in my example "If" statement means you don't have to store the random number in a variable at all. It just executes whatever is inside the "If" about 20% of the time (on my single test run it was 18%).

VBAjedi

 

[haytatreides]

i got it. the cint doesn't work no need for it. thanks all