Format string for usa phone number in vb.net

Question

as the question says. I get numbers like 2125550938 or 20298277625552. They should change to [phone] and [phone] x 5552 respectively. it is in vb.net

Answer 1

Try the following when using String.Format in VB.NET:

String.Format("{0:(###) ###-####}", Long.Parse(PhoneString)) 

Answer 2

I would probably go with the regex implementation, something like this:

  Dim phoneNumbers() As String = {"2125550938", _ "20298277625552", _ "2025551212378", _ "202555131345943"} Dim ext As String = "" Dim r As New Regex("^(?<AC>\d{3})(?<First>\d{3})(?<Last>\d{4})(?<Ext>\d*$)") Dim m As Match For i As Int32 = 0 To (phoneNumbers.Length - 1) m = r.Match(phoneNumbers(i)) If m.Groups("Ext").Length > 0 Then ext = " x " & CStr(m.Groups("Ext").Value) Else ext = "" End If Console.WriteLine("({0}) {1}-{2}{3}", _ CStr(m.Groups("AC").Value), _ CStr(m.Groups("First").Value), _ CStr(m.Groups("Last").Value), ext) Next Console.Read() 

This will allow you to use phone numbers without extensions or with variable length extensions.

Answer 3

Open Common Phone PhoneFormat (ByVal strPhoneNumber As String) As String

 ' Remove any style characters from the user input strPhoneNumber = Replace(strPhoneNumber, ")", "") strPhoneNumber = Replace(strPhoneNumber, "(", "") strPhoneNumber = Replace(strPhoneNumber, "-", "") strPhoneNumber = Replace(strPhoneNumber, ".", "") strPhoneNumber = Replace(strPhoneNumber, Space(1), "") Dim strFormatedNumber As String = CLng(strPhoneNumber).ToString("(###) ###-####") Return strFormatedNumber 

Final function

Answer 4

Here is a combination of regular and string formatting that works well enough for me. Just add salt to the recipe and it will be delicious!

 ''' <summary> ''' returned formats are: ''' example1 [phone] Ext: 890 ''' example2 123-4567 Ext: 890 ''' example3 [phone] ''' example4 123-4567 ''' The user can input a 7 or 10 digit number followed by a character(s) and digits for the extension ''' ''' </summary> ''' <param name="OriginalNumber"></param> ''' <returns></returns> ''' <remarks></remarks> Public Function FormatPhone(ByVal OriginalNumber As String) As String Dim sReturn As String Dim tester() As String Dim R As Regex Dim M As Match Dim sTemp As String sReturn = "" ' removes anything that is not a digit or letter sTemp = UnFormatPhone(OriginalNumber) ' splits sTemp based on user input of character(s) to signify an extension ie x or ext or anything else you can think of (abcdefg...) tester() = Regex.Split(sTemp, "\D+") ' if the string was split then replace sTemp with the first part, ie the phone number less the extension If tester.Count > 1 Then sTemp = tester(0) End If ' Based on the NANP (North American Numbering Plan), we better have a 7 or 10 digit number. anything else will not parse If sTemp.Length = 7 Then R = New Regex("^(?<First>\d{3})(?<Last>\d{4})") ElseIf sTemp.Length = 10 Then R = New Regex("^(?<AC>\d{3})(?<First>\d{3})(?<Last>\d{4})") Else Return OriginalNumber End If ' now format the phone number nice and purtee... M = R.Match(sTemp) If m.Groups("AC").Length > 0 Then sReturn &= String.Format("({0}) {1}-{2}", CStr(m.Groups("AC").Value), CStr(m.Groups("First").Value), CStr(m.Groups("Last").Value)) Else sReturn &= String.Format("{0}-{1}", CStr(m.Groups("First").Value), CStr(m.Groups("Last").Value)) End If If tester.Count > 1 Then sReturn &= " Ext: " + tester(1) End If Return sReturn End Function ''' <summary> ''' Strips NON ALPHANUMERICS from a string ''' ''' </summary> ''' <param name="sTemp"></param> ''' <returns></returns> ''' <remarks></remarks> Public Function UnFormatPhone(ByVal sTemp As String) As String sTemp = ClearNull(sTemp) Dim sb As New System.Text.StringBuilder For Each ch As Char In sTemp If Char.IsLetterOrDigit(ch) OrElse ch = " "c Then sb.Append(ch) End If Next UnFormatPhone = sb.ToString End Function ''' <summary> ''' Returns a trimmed string with vbNullChar replaced by a blank ''' </summary> ''' <param name="sTemp"></param> ''' <returns></returns> ''' <remarks></remarks> Public Function ClearNull(ByRef sTemp As String) As String sTemp = Replace(sTemp, vbNullChar, "") ClearNull = Trim(sTemp) End Function 

Answer 5

 Dim newNumber As New String If number.Length = 10 Then newNumber = "(" & number.Substring(0, 3) & ") " & number.Substring(2, 3) & "-" & number.Substring(5, 4) ElseIf number.Length = 14 Then newNumber = "(" & number.Substring(0, 3) & ") " & number.Substring(2, 3) & "-" & number.Substring(5, 4) & " x " & number.Substring(9) End if