Problem with getting word count in TLF text

Question

Hi,
I want to get the word count from my TLF text, but the problem is that I am not being able to handle th case for space.
I am using the findNextWordBoundary property of ParagraphElement as shown below:
private function countWords( para : ParagraphElement ) : void
            var wordBoundary:int = 0;
            var prevBoundary:int = 0;
            while ( wordBoundary != para.findNextWordBoundary( wordBoundary ) )
               // If the value is greater than 1, then it's a word, otherwise it's a space.
                if ( para.findNextWordBoundary( wordBoundary ) - wordBoundary > 1)
                    wordCount += 1;                   
                prevBoundary = wordBoundary;
                wordBoundary = para.findNextWordBoundary( wordBoundary );                   
                // If the value is greater than 1, then it's a word, otherwise it's a space.
                if ( wordBoundary - prevBoundary > 1 )
                    var s:String = para.getText().substring( prevBoundary, wordBoundary );
                    lenTotal += s.length;
Now I have 2 issues here:
If my string is for eg: Hi, I am writing in "TLF". And I want to get its word count then
1) Suppose I take the case of the string Hi,  . Then para.getText().substring( prevBoundary, wordBoundary ) gives the text as Hi i.e without the comma. Same case for the string "TLF forums" , It treats each " as a single word and not the whole "TLF" as a single word. Why doesn't it compute till spaces, that should be the ideal case. So until we don't give a space it should count the whole thing as a word.
2) So now the problem is I have applied a condition   if ( wordBoundary - prevBoundary > 1 ) to check if it is a space i.e. if the diff is <= 1 it is a Space. But if I use this I miss out on single words. Like for eg if I have "Hi, This is a string" ,then 'a' is ignored too.
Now I could have added a check here along with the space check that the string between prevBoundary and wordBoundary is " "(i.e a space), Then also it is a problem as then the single words like a,&,I will be ignored.
So, now I am stuck with this issue and need some help from you guys.
Thanks

Answer 1

findNextWordBoundary is not going to serve your purpose.  I'd propose doing something like this:
// didn't test this but something like this - whitespace matches any set of 1 or more white space characters
static const whiteSpaceRegExp:RegExp = /[u0020|u000A|u000D]*/
public static function countWords( para : ParagraphElement ) : void
     return para.getText().split(whiteSpaceRegExp).length;
A good list of everything considered whitespace extracted from the unicode space can be found here:
http://sourceforge.net/adobe/tlf/svn/449/tree/trunk/textLayout/src/flashx/textLayout/utils /CharacterUtil.as
In function createWhiteSpaceObject
Hope that helps,
Richard