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Posted By

Saikou on 11/22/07


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sbbath


Name extractor

Published in: Visual Basic 

Attempts to extract a first name and last name from a string, such as "Mr Smith", "Smith, John", "John James Smith", "J Smith", "Smith, J" etc.

Plain TextExpand 
  1. Private Sub ParseName(ByVal strInput, ByRef strFirstName, ByRef strLastName)
  2.         Try
  3.             ' Split the string into it's constituent words
  4.             Dim strSplitName1() As String = strInput.Split(New Char() {" "})
  5.  
  6.             ' Ensure any leading or trailing spaces are trimmed within each word
  7.             Dim intIndex As Integer = 0
  8.             For intIndex = 0 To strSplitName1.Length - 1
  9.                 strSplitName1(intIndex) = strSplitName1(intIndex).Trim
  10.             Next
  11.  
  12.             ' If only one word was found, treat it as a last name, with no first name
  13.             If (strSplitName1.Length <= 1) Then
  14.                 strFirstName = ""
  15.                 strLastName = strSplitName1(0)
  16.                 Return
  17.             End If
  18.  
  19.             ' Check the first word for a title
  20.             If ( _
  21.                 (String.Compare(strSplitName1(0), "Mr", True) = 0) OrElse _
  22.                 (String.Compare(strSplitName1(0), "Mr.", True) = 0) OrElse _
  23.                 (String.Compare(strSplitName1(0), "Mister", True) = 0) OrElse _
  24.                 (String.Compare(strSplitName1(0), "Mrs", True) = 0) OrElse _
  25.                 (String.Compare(strSplitName1(0), "Ms", True) = 0) OrElse _
  26.                 (String.Compare(strSplitName1(0), "Ms.", True) = 0) OrElse _
  27.                 (String.Compare(strSplitName1(0), "Miss", True) = 0) OrElse _
  28.                 (String.Compare(strSplitName1(0), "Dr", True) = 0) OrElse _
  29.                 (String.Compare(strSplitName1(0), "Dr.", True) = 0) OrElse _
  30.                 (String.Compare(strSplitName1(0), "Sir", True) = 0) OrElse _
  31.                 (String.Compare(strSplitName1(0), "Professor", True) = 0) OrElse _
  32.                 (String.Compare(strSplitName1(0), "Prof", True) = 0) OrElse _
  33.                 (String.Compare(strSplitName1(0), "Lord", True) = 0) OrElse _
  34.                 (String.Compare(strSplitName1(0), "Lady", True) = 0) OrElse _
  35.                 (String.Compare(strSplitName1(0), "Rev", True) = 0) OrElse _
  36.                 (String.Compare(strSplitName1(0), "Rev.", True) = 0) OrElse _
  37.                 (String.Compare(strSplitName1(0), "Reverend", True) = 0) OrElse _
  38.                 (String.Compare(strSplitName1(0), "Judge", True) = 0) OrElse _
  39.                 False _
  40.                 ) Then
  41.  
  42.                 ' A title was found, so the second and last words should be first and last name respectively
  43.                 If (strSplitName1.Length > 2) Then
  44.                     strFirstName = strSplitName1(1)
  45.                     strLastName = strSplitName1(strSplitName1.Length - 1)
  46.                 Else
  47.                     ' Only a title and one other word was found, so treat the other word as a last name
  48.                     strLastName = strSplitName1(1)
  49.                 End If
  50.             ElseIf (strSplitName1(0).Length = 1) Then
  51.                 ' The first word is likely an initial, so use the first non-length-of-1 word as the last name
  52.                 strFirstName = strSplitName1(0)
  53.                 For Each strWord As String In strSplitName1
  54.                     If (strWord.Length > 1) Then
  55.                         strLastName = strWord
  56.                     End If
  57.                 Next
  58.             ElseIf (strSplitName1(0).Contains(",")) Then
  59.                 ' Check the first word for a comma, if it contains one, it's likely to be 'Last, First' format
  60.                 strFirstName = strSplitName1(0).Replace(",", "").Trim()
  61.                 strLastName = strSplitName1(1)
  62.             Else
  63.                 ' It's most likely to be a "First (Middle(s)) Last" format, so simply use the first and last words
  64.                 strFirstName = strSplitName1(0)
  65.                 strLastName = strSplitName1(strSplitName1.Length - 1)
  66.             End If
  67.         Catch ex As Exception
  68.             'EventLogger.LogException(ex)
  69.         End Try
  70.     End Sub

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