Counting Characters less whitespace, with Regular Expression
VB.NET 2003
I'm trying to count the words and characters in a text box. I picked up the CountWords function from a post by TheLearnedOne and it works great. I was trying to build off of it and make a function that would count all characters with the exception of any whiteSpace. The function I wrote, CountCharacters is not working. Any help in counting the characters without counting spaces would really be appreciated.
Public Shared Function CountWords(ByVal inputText As String) As Integer
Dim patternWords As String = "[\w]+"
Dim regCountWords As New System.Text.RegularExpress
Return regCountWords.Matches(inpu
End Function 'CountWords'
Public Shared Function CountCharacters(ByVal inputText As String) As Integer
Dim patternCharacters As String = "[^:Wh]"
Dim regCountCharacters As New System.Text.RegularExpress
Return regCountCharacters.Matches
End Function 'CountCharacters'
I'm trying to count the words and characters in a text box. I picked up the CountWords function from a post by TheLearnedOne and it works great. I was trying to build off of it and make a function that would count all characters with the exception of any whiteSpace. The function I wrote, CountCharacters is not working. Any help in counting the characters without counting spaces would really be appreciated.
Public Shared Function CountWords(ByVal inputText As String) As Integer
Dim patternWords As String = "[\w]+"
Dim regCountWords As New System.Text.RegularExpress
Return regCountWords.Matches(inpu
End Function 'CountWords'
Public Shared Function CountCharacters(ByVal inputText As String) As Integer
Dim patternCharacters As String = "[^:Wh]"
Dim regCountCharacters As New System.Text.RegularExpress
Return regCountCharacters.Matches
End Function 'CountCharacters'
ASKER
That works but I still have a problem. The CountCharacters function is not counting spaces now, but it also is not counting any puncuation.
Looking up Regular Expressions in help I see that:
[^...] = Matches any character not in the set of characters following the ^.
:Wh = Matches all types of whitespace, including publishing and ideographic spaces.
This is where I though that
Dim patternCharacters As String = "[^:Wh]"
would count any characters except whitespace.
Does anyone know the correct Regular Expressions syntax that will count all characters except whitespace?
Looking up Regular Expressions in help I see that:
[^...] = Matches any character not in the set of characters following the ^.
:Wh = Matches all types of whitespace, including publishing and ideographic spaces.
This is where I though that
Dim patternCharacters As String = "[^:Wh]"
would count any characters except whitespace.
Does anyone know the correct Regular Expressions syntax that will count all characters except whitespace?
ASKER CERTIFIED SOLUTION
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ASKER
I think we have it. I had to make a few changes. I realized that the "[^:Wh ]" was removing the "h" as well as the "whiteSpace". Then I realized I needed to remove the "newline" and "tab". I know there are other ways to do this but I am using this in real time as the operator types the counts are continously updated. That's why I wanted to go with the Regular Expressions because of the speed. Thanks for the help.
Public Shared Function CountCharacters(ByVal inputText As String) As Integer
Dim patternCharacters As String = "[^:W \n \t ]" 'whiteSpace, newline, and tabs removed from count
Dim regCountCharacters As New System.Text.RegularExpress
Return regCountCharacters.Matches
End Function 'CountCharacters'
Public Shared Function CountCharacters(ByVal inputText As String) As Integer
Dim patternCharacters As String = "[^:W \n \t ]" 'whiteSpace, newline, and tabs removed from count
Dim regCountCharacters As New System.Text.RegularExpress
Return regCountCharacters.Matches
End Function 'CountCharacters'
You can try just removing the spaces first then counting individual characters like this;
Public Function CountCharacters(ByVal inputText As String) As Integer
Dim tempText As String = inputText.Replace(" ", "")
Dim patternCharacters As String = "[\w]"
Dim regCountCharacters As New System.Text.RegularExpress
Return regCountCharacters.Matches
End Function 'CountCharacters'
Hope this helps,
Neil