Removing a character

[Moggie webforumsuser@macromedia.com]

Probably very easy, but I'm stumped!!

I have this:

on keyDown me
legalCharacters = "abcdefghijklmnopqrstuvwxyz"
if legalCharacters contains the key then
put "*" after member "star"


to create a * for each letter typed. How can I get member "star" to decrease the number of * if the user presses backspace?

Morag

[Rob Dillon]

Make these few changes and you can use the backspace/delete key:

> on keyDown me
> legalCharacters = "abcdefghijklmnopqrstuvwxyz"

thisManyChars = member("star").char.count
if the keyCode = 51 and thisManyChars > 0 then
member("star").char[thisManyChars].delete()
end if

> if legalCharacters contains the key then
> put "*" after member "star"

end if
end

--
Rob
_______
Rob Dillon
Team Macromedia
[url]http://www.ddg-designs.com[/url]
412-243-9119

[url]http://www.macromedia.com/software/trial/[/url]

[Gretchen webforumsuser@macromedia.com]

<<
on keyDown me
legalCharacters = "abcdefghijklmnopqrstuvwxyz"
if legalCharacters contains the key then
put "*" after member "star"


to create a * for each letter typed. How can I get member "star" to decrease the number of * if the user presses backspace?>>

legalCharacters = "abcdefghijklmnopqrstuvwxyz"
if legalCharacters contains the key then
put "*" after member "star"
else if charToNum(the key) = 8 then
str = member("star").text
delete the last char of str
member("star").text = str
end if

Gretchen Macdowall
updateStage, inc.

[Moggie webforumsuser@macromedia.com]

Thank you, Gretchen. It works perfectly.

Morag