Checkbox javascript problem
lazarus / 38-Networking and Web Programming
----------------------------------------------------------------------
_Julia_ · 2015-01-24 17:29
Hello,
i'm making this code and i m trying to create code that can count the
checkboxes together.
for now i have writen this code : not important verbs are in my
language so ignore that ..
i m looking for any ideas how can i change this code to be able to
count both checkboxes for now i can only pick one and when i do both
my reasults came wrong ... it counts only 1.
so i would be glad if anyone could help me with that.
so my question is how to rewrite the code to be able to count both
checkboxes into the result. thanks { if there are more then 2 then ofc
it have to count all of them. }
A simple frameset document
Dan z prijmu fyzickych osob
----------------------------------------------------------------------
skalogryz · 2015-01-24 18:49
instead of assigning the value to "z" variable, you should increase
its value, just replace
Code: [Select] z=typ.value;
with
Code: [Select] z+=typ.value;
----------------------------------------------------------------------
_Julia_ · 2015-01-27 21:28
well i tried but the code it's still not working right the same
problem that it counts only the 1 possibility.
i tryed everything but i cant find anything in pure javascript that do
that ... blew
----------------------------------------------------------------------
TurboRascal · 2015-01-27 22:00
Isn't this forum a strange place to come for help about javascript?
----------------------------------------------------------------------
skalogryz · 2015-01-27 22:19
Quote from: _Julia_ on January 27, 2015, 10:28:50 pmwell i tried but
the code it's still not working right the same problem that it counts
only the 1 possibility.
Your loop is correct, but elements of "typ"-array are misused.
This is how it should be.
Code: [Select] var z = 0;
for (var i = 0; i < typ.length; i++) {
if (typ[i].type === 'checkbox' && typ[i].checked) {
z += parseInt(typ[i].value);
}
note: that value is now wrapped into parseInt. Otherwise you'd see an
unexpected 1050, instead of 60 as value of "z"
----------------------------------------------------------------------
skalogryz · 2015-01-27 22:20
Quote from: TurboRascal on January 27, 2015, 11:00:15 pmIsn't this
forum a strange place to come for help about javascript?
well... isn't it "Programming / Network and Web Programming" sub
section ?