Thread: VBA conundrum
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Bob Phillips
 
Posts: n/a
Default VBA conundrum

It declares a variable, called myVal, of type Long, which means that it can
hold an integer value between -2,147,483,648 and 2,147,483,647. It is this
variabel that I store the user input number.

--

HTH

RP
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"csi" wrote in message
...
Thank-you for the info; it was quite useful.
By the way, what does "Dim myVal As Long" do in terms of this macro?
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vbab


"Bob Phillips" wrote:

Outdoing yourself for helpfulness today, aren't you.

OP,

Here is a little macro that does what I think you ask. It doesn't get

column
C as you predict, so it probably means I don't understand what you need
there.

Dim myVal As Long
Const nDefault As Long = 6000
Const nMin As Long = 20000
Const nMax As Long = 30000

Application.ScreenUpdating = False
myVal = InputBox("Input a number between 0 and 10,000")
If myVal nMin And myVal < nMax Then
If myVal nDefault Then
For i = 1 To myVal \ nDefault
Cells(i, "A").Value = i
Cells(i, "B").Value = nDefault
Cells(i, "C").Value = (nDefault / 1000) * (nDefault /

1000)
* 15
Next i
End If
Cells(i, "A").Value = i
Cells(i, "B").Value = myVal Mod nDefault
Cells(i, "C").Value = (Cells(i, "B").Value / 1000) * _
(Cells(i, "B").Value / 1000) * 15
End If
Application.ScreenUpdating = True

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HTH

RP
(remove nothere from the email address if mailing direct)


"Jezebel" wrote in message
...

Is it possible to do all these calculations easily?


Yes.