Forums Scripts & Popups regular expression - calc MAIUSC percent

Hi I have this code:


It tell me only if someone write in MAIUSC. But, if I would like to calc how many words are in MAIUSC in the string?

$regex($1-,/\b[A-Z]+\b/g)

Note that right now the boundaries for the uppercase alpha words are \b, which will match the position between a non alphanumeric char and an alphanumeric char.

I do this so that in the sentence: WHAT? It would say there's 1 word in upper case, even though technically the string "WHAT?" is not an alpha string, since it contains a question mark. Also in a string like: HEY, you! it will also say there's 1 word in upper case, even though the string "HEY," isn't an alpha string.

Consequence of this is that on the word "SEMI-advanced" it will say there is 1 word in upper case, even though the entire word contains an "-" sign, so isn't technically an alpha char.

If you do only want it to match nothing but alpha chars that can only be surrounded by either a space or the start/end of the string, you could specify:

$regex($1-,/(?<= |^)[A-Z]+(?= |$)/g)


Note that you didn't need a regex for this, a simple loop like the following would have worked just fine.

var %a
while ($0) {
if ($1 isupper) && ($1 isalpha) inc %a
tokenize 32 $2-
}
echo -a %a

yes, but this not count how many words are in uppercase. I would like to count how many words are in uppercase in the string.

It does. Test it before posting.

I have tested it and it show only the word without the ? character.

//tokenize 32 IT DOES WORK | echo -a $regex($1-,/(?<= |^)[A-Z]+(?= |$)/g)

//tokenize 32 IT DOES WORK | echo -a $regex($1-,/\b[A-Z]+\b/g)



//echo -a $upperwords(IT DOES WORK)

ok, Thanks for the help

Sorry if I am too stupid