[RESOLVED] Making InputBox longer?

[BubbleLife]

Hi all,

Just a quick question - is it possible to force the size of inputbox to be longer, without having to make my own one as a form? To stop problems like this:



I tried changing the XPos and YPos, but that's just the position of the inputbox on the screen....

Thanks!

[thingimijig]

Split the number up into lines

InputBox "Please input the following number" _
         & vbCrLf & "123123123123123123123123123123123" _
         & vbCrLf & "123123123123123123123123123123123" _
         & vbCrLf & "123123123123123123123123123123123"

[BubbleLife]

Ah, but the number is randomly generated . The actual code is:

IntakeUserNumber = InputBox("Please input the following number: " & TotNumberInput)

Maybe there's a way of cutting down the string into, say, 25-character chunks and displaying them on different lines?

[gavio]

Here's a quick workaround:

Dim tmp1 As String
Dim tmp2 As String
 
tmp1 = "3498302938349830293834983029383498" & _
       "302938349830293834983029383498302" & _
       "938349830293834983029383498302938" '100 characters
 
tmp2 = InputBox(Left(tmp1, 34) & vbCrLf & Mid(tmp1, 33, 33) & vbCrLf & Right(tmp1, 33), "Is it OK?", "0")

[BubbleLife]

How would I do that when I don't know the number, though? (as in my last post)

[Comintern]

How about a small function to wrap long lines:

IntakeUserNumber = InputBox(SplitString("Please input the following number: " & TotNumberInput))
 
Private Function SplitString(ByVal sString As String, Optional lLen As Long = 25) As String
 
    Dim lCount As Long
    
    If Len(sString) <> 0 Then
        For lCount = 1 To Len(sString) \ lLen
            SplitString = SplitString & Left$(sString, lLen) & vbCrLf
            sString = Right$(sString, Len(sString) - lLen)
        Next lCount
 
        SplitString = SplitString & sString
    End If
 
End Function

[BubbleLife]

Works perfect! If only I could understand the code, heh.

Thanks a lot

[Comintern]

Works perfect! If only I could understand the code, heh.

Thanks a lot

Commented version below

Private Function SplitString(ByVal sString As String, Optional lLen As Long = 25) As String
 
    Dim lCount As Long
    
    If Len(sString) <> 0 Then                           'Check for vbNullString (avoids divide by zero).
        For lCount = 1 To Len(sString) \ lLen           'The math determines the number of full chunks (int divide)
                                                        'and then loops that many times.
            SplitString = SplitString & Left$(sString, lLen) & vbCrLf   'Add the current chunk to the output string.
            sString = Right$(sString, Len(sString) - lLen)  'Remove it from the input string.
        Next lCount
 
        SplitString = SplitString & sString             'If there's anything left in the input, tack it on the output.
    End If
 
End Function

[BubbleLife]

Ahh.. Confusing, haha. Well, I'm only a VB beginner (everything I know about VB I learnt from a short unit in my BTEC...), and I suck at maths, soo.. heh. But thank you for commenting it anyway