\r\n \r\n Phil\r\n \r\n

Hi,
\nJust an initial thought, you could use the "Split" function with a space as the deliminator to return an array of the words, then the ubound of the array + 1 would give the number of words.
\n
\n Dim Words() As String
\n Dim NoWords As Integer
\n
\n Words = Split(Trim(Text1.Text), " ")
\n NoWords = UBound(Words) + 1
\n
\nIncluding Trim ensures leading or trailing spaces are not counted. There may be a better way but this is my initial thought, Good Luck
\nPhil
\n
\n
\n

\r\n \r\n qwertyjustin\r\n \r\n

didn\'t seem to work, any simpler method?(and i need it in a label, the amount
\n
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\r\n \r\n Phil\r\n \r\n

Ah,
\nJust thought of it what version of vb are you using.
\nSplit is only available in VB6, there is a work round however which is detailed in MSDN article Q188007,
\n
\nhttp://support.microsoft.com/support...S&SD=msdn&FR=0
\n
\nyou can copy a function which will do the same job as Split.
\n
\nTrying to think of a simpler method, get back to you if I can.
\nCheers
\nPhil
\n
\n

\r\n \r\n Eagle\r\n \r\n

I don\'t know if this is any more simple but you could always use a loop to cycle through each character (using SelStart, SelText, and SelLength to select each character and test them one at a time) and everytime you hit a character that is a space add 1 to a counter. Then when the you get to the end of a string add 1 to the counter for the last word. I recently had to remove a file name from its path and I did this same thing, only I counted the """ marks and when I reached the last """ then I knew that was the begining of the file name. If this is confusing I might be able to clear it up with some example code. Let me know.
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\r\n \r\n BillSoo\r\n \r\n\r\n\r\n