Function ParseWord(varPhrase As Variant, ByVal iWordNum As Integer, Optional strDelimiter As String = " ", _
Optional bRemoveLeadingDelimiters As Boolean, Optional bIgnoreDoubleDelimiters As Boolean) As Variant
On Error GoTo Err_Handler
'Purpose: Return the iWordNum-th word from a phrase.
'Return: The word, or Null if not found.
'Arguments: varPhrase = the phrase to search.
' iWordNum = 1 for first word, 2 for second, ...
' Negative values for words form the right: -1 = last word; -2 = second last word, ...
' (Entire phrase returned if iWordNum is zero.)
' strDelimiter = the separator between words. Defaults to a space.
' bRemoveLeadingDelimiters: If True, leading delimiters are stripped.
' Otherwise the first word is returned as null.
' bIgnoreDoubleDelimiters: If true, double-spaces are treated as one space.
' Otherwise the word between spaces is returned as null.
'Author: Allen Browne. [URL unfurl="true"]http://allenbrowne.com.[/URL] June 2006.
Dim varArray As Variant 'The phrase is parsed into a variant array.
Dim strPhrase As String 'varPhrase converted to a string.
Dim strResult As String 'The result to be returned.
Dim lngLen As Long 'Length of the string.
Dim lngLenDelimiter As Long 'Length of the delimiter.
Dim bCancel As Boolean 'Flag to cancel this operation.
Const CHARS = ".!?,;:""'()[]{}#"
Dim intIndex As Integer
'*************************************
'Validate the arguments
'*************************************
'Cancel if the phrase (a variant) is error, null, or a zero-length string.
If IsError(varPhrase) Then
bCancel = True
Else
strPhrase = Nz(varPhrase, vbNullString)
If strPhrase = vbNullString Then
bCancel = True
End If
End If
'If word number is zero, return the whole thing and quit processing.
If iWordNum = 0 And Not bCancel Then
strResult = strPhrase
bCancel = True
End If
'Delimiter cannot be zero-length.
If Not bCancel Then
lngLenDelimiter = Len(strDelimiter)
If lngLenDelimiter = 0& Then
bCancel = True
End If
End If
'*************************************
'Process the string
'*************************************
If Not bCancel Then
strPhrase = varPhrase
'Remove leading delimiters?
If bRemoveLeadingDelimiters Then
strPhrase = Nz(varPhrase, vbNullString)
Do While Left$(strPhrase, lngLenDelimiter) = strDelimiter
strPhrase = Mid(strPhrase, lngLenDelimiter + 1&)
Loop
End If
'Remove Chars
For intIndex = 1 To Len(CHARS)
strPhrase = Trim(Replace(strPhrase, _
Mid(CHARS, intIndex, 1), " "))
Next intIndex
'Ignore doubled-up delimiters?
If bIgnoreDoubleDelimiters Then
Do
lngLen = Len(strPhrase)
strPhrase = Replace(strPhrase, strDelimiter & strDelimiter, strDelimiter)
Loop Until Len(strPhrase) = lngLen
End If
'Cancel if there's no phrase left to work with
If Len(strPhrase) = 0& Then
bCancel = True
End If
End If
'*************************************
'Parse the word from the string.
'*************************************
If Not bCancel Then
varArray = Split(strPhrase, strDelimiter)
If UBound(varArray) >= 0 Then
If iWordNum > 0 Then 'Positive: count words from the left.
iWordNum = iWordNum - 1 'Adjust for zero-based array.
If iWordNum <= UBound(varArray) Then
strResult = varArray(iWordNum)
End If
Else 'Negative: count words from the right.
iWordNum = UBound(varArray) + iWordNum + 1
If iWordNum >= 0 Then
strResult = varArray(iWordNum)
End If
End If
End If
End If
'*************************************
'Return the result, or a null if it is a zero-length string.
'*************************************
If strResult <> vbNullString Then
ParseWord = strResult
Else
ParseWord = Null
End If
Exit_Handler:
Exit Function
Err_Handler:
Call LogError(Err.Number, Err.Description, "ParseWord()")
Resume Exit_Handler
End Function