Help with text mask

What are the allowed patterns? Can you post some examples?

sry Erel,
for currency i need: 9999.99
and for date: dd/mm/yy

I don't recommend you to use TextChanged event to check that the date string is valid. The user experience will be very bad.
You should instead validate the string when you do something with it.

To only allow decimal numbers you should use:

The easiest way to validate the date is to try and parse it and use Try Catch to catch the error if it is not valid.

Hi Erel,
I'm doing a program for collectors of money for companies that use our system (windows), and this is kind of hard headed people ...
I need to give all chewed...

I make the check date with the TextChanged, but i'cant erase the text inside the edittext... =(

Can you help me?

As I said in my previous post, doing this check in TextChanged will not work properly. You should assume that at some points the string will not be a valid date.

hi,
see below
demo for mobile fone mask


Sub edtCelular_TextChanged (Old As String, New As String)
Dim current As String
current = New

Dim positioning(12) As Int
'positioning(1) = 1, 2, 3, 6, 7, 8, 9, 11, 12, 13, 14 ) As Int
'(99) 1234-1234
positioning(0) = 1
positioning(1) = 2 '1
positioning(2) = 3 '2
positioning(3) = 6 ' 3
positioning(4) = 7
positioning(5) = 8
positioning(6) = 9
positioning(7) = 11
positioning(8) = 12
positioning(9) = 13
positioning(10) = 14

If (isUpdating) Then
isUpdating = False
Return
End If

Dim number As String

number=""
Dim letra As String
Dim lista As String
lista="0123456789"
For i=0 To current.length-1
letra=current.CharAt(i)
If lista.IndexOf(letra) >=0 Then
number=number & letra
End If


Next
Dim maxNumberLength As Int
maxNumberLength=11

If (number.length > 10) Then
number = number.substring2(0, 10)
'number.sub
End If
Dim length As Int
length = number.length

' Pad the number To 10 characters...
Dim paddedNumber As String
paddedNumber = padNumber(number, maxNumberLength)
' Split phone number into parts... /
Dim ddd,part1,part2 As String
Try
ddd = paddedNumber.substring2(0, 2)
Catch
End Try

Try
part1 = paddedNumber.substring2(2, 6)
Catch
End Try

Try
part2 = paddedNumber.substring2(6, 10)
Catch
End Try
' build the masked phone number... */
Dim Phone As String

Phone = "(" & ddd & ") " & part1 & "-" & part2

isUpdating = True
edtCelular.Text=Phone
edtCelular.SelectionStart = positioning( length )
End Sub

Sub padNumber( number As String, maxLength As Int) As String
'String padded = new String(number);
Dim padded As String
Dim i As Int
padded=number
For i = 0 To maxLength - number.length
padded = padded & " "
Next
Return padded

End Sub

amigo, se precisar tenho outras mascaras
sou brasileiro,ok (31) 94136747