Count characters in a String

[wanderallyouwant]

Looking for something like this:

Func _LettersInString($string)
Dim $result, $i
;<not sure> of the following
$letters = StringSplit($string)
for $i = 0 to UBound($letters)

if IsCharacter($letters[$i]) then $result = $result + 1


Next

;</not sure>
Return $result
EndFunc


$sample = "Hey, what's for lunch at 12?"
MsgBox(0,"","There are " & _LettersInString($sample) & " letters in the string.")

Edited July 20, 2007 by wanderallyouwant

[xcal]

StringLen. And use StringStripWS if you don't want the spaces.

[wanderallyouwant]

But I only want letters, like not a-z. Would my above code work if my IsCharacter func existed, or is there already one?

[wanderallyouwant]

I guess the appropriate title should have been Count letters in a String rather than characters.

[xcal]

$str = 'Hey, what''s for lunch at 12?'
$ret = StringRegExp($str, '[^\d\W]', 3)

MsgBox(0, '', UBound($ret))

edit2 - undid my original edit, since I had it right in the first place

Changed the pattern to be more consistant, too.

Maybe a regexp expert could tell me why [^\d\W] works, but [\D\w] doesn't? They should be the same, no?

Edited July 20, 2007 by xcal

[SpookMeister]

I believe UBound is only for arrays

Instead of UBound, use StringLen in your script example like xcal mentioned above

[xcal]

You realize it's returning an array for each letter, right?

[SpookMeister]

Hey... gimme a break.. I just woke up

This might work for him too

[edit.. realized he wanted a numerical return]

$string = "Happy 16th Birthday"
MsgBox(4096,"",$string & @CRLF & _LettersInString($string))
Func _LettersInString($sString)

    $output = ""
    For $i = 1 To StringLen($sString)
        $char = StringMid($sString, $i, 1)
        If Not Number($char) Then $output = $output & $char
    Next
    $output = StringStripWS($output, 8)
    Return StringLen($output)
EndFunc   ;==>_LettersInString

Edited July 20, 2007 by SpookMeister

[xcal]

Or you could also do...

$str = 'Hey, what''s for lunch at 12?'
$ret = StringRegExpReplace($str, '[\d\W]', '')

MsgBox(0, '', StringLen($ret))

; edit - Or this probably makes the most sense...

$str = 'Hey, what''s for lunch at 12?'
$ret = StringRegExp($str, '(?i)[a-z]', 3)

MsgBox(0, '', UBound($ret))oÝ÷ ØéÚ­éZµçg¢×­è^âëÊÍ=Û"Z®¢ÙµÈZ*®¢Ü(­Ú®¢×!j¶µêÚÌh®éÝz»¢·ªº[ºÛay·¥£hzÉ÷öÜ(®F®¶­sbb33c·7G"Òb33´WÂvBb33²b33·2f÷"ÇVæ6B#òb33°¢b33c·&WBÒ7G&æu&VtWb33c·7G"Âb33²b3#·rb33²Â2 ¤×6t&÷Âb33²b33²ÂT&÷VæBb33c·&WB

It also returns the 1 and the 2. Did I misunderstand something?

Edited July 20, 2007 by xcal

[Siao]

$sample = "Hey, what's for lunch at 12?"
MsgBox(0,"","There are " & _LettersInString($sample) & " letters in the string.")

Func _LettersInString($string)
    Return StringLen(StringRegExpReplace($string, '[^[:alpha:]]', ''))
EndFunc

Edited July 20, 2007 by Siao

[wanderallyouwant]

$sample = "Hey, what's for lunch at 12?"
MsgBox(0,"","There are " & _LettersInString($sample) & " letters in the string.")

Func _LettersInString($string)
    Return StringLen(StringRegExpReplace($string, '[^[:alpha:]]', ''))
EndFunc

Works great, how can I make it work for numbers? I tried changing alpha to numeric, numeral and all forms of the word I could think of.

[Paulie]

Works great, how can I make it work for numbers? I tried changing alpha to numeric, numeral and all forms of the word I could think of.

Here is how you would do for Numbers

$sample = "So the 2 movies start at 1:00, 2:00, and 4:00?"
MsgBox(0,$Sample,"There are " & _NumbersInString($sample) & " numbers in the string.")

Func _NumbersInString($string)
    Return StringLen(StringRegExpReplace($string, '[^[:digit:]]', ''))
EndFunc

The helpfile is like Prego. Its in there

Edited July 21, 2007 by Paulie