Javascript not working with adding variables together

Question

Right so i have values in 3 dropdown boxes and there will be more (looking at adding about 20 dropdown boxes that add to give a final price for the item. Here is the code for the HTML.

<form id="myform1>
<p>Testing 1</p>
<select id="Test1">
<option value="24">Item one</option>
<option value="29">Item Two<option>
<!--I have the first box going to 20 items-->
</select>

<p>Testing 2</p>
<select id="Test2">
<option value="25">Item one</option>
<option value="100">Item Two<option>
<!--I have the second box going to 10 items-->
</select>

<p>Testing 3</p>
<select id="Test3">
<option value="24">Item one</option>
<option value="100">Item Two<option>
<!--I have the third box going to 10 items-->
</select>


</form>
<button onclick="myFunction()">Update</button>
<p id="demo"></p>

Then i have my Javascript

function myFunction()
{
var a = document.getElementById("list1");
var A = a.options[a.selectedIndex].value;

var b = document.getElementById("list2");
var B = b.options[a.selectedIndex].value;

var c = document.getElementById("list3");
var C = c.options[a.selectedIndex].value;

var IntA = parseInt(A);
var IntB = parseInt(B);
var IntC = parseInt(C);

var x=IntB + IntA;
var y=x + IntC;
var demoP=document.getElementById("demo")
demoP.innerHTML= y;
}

My Problem is it only works if i take var c our completely and then take var y out and have it calculate to x. I have tryed many ways and am at a loss. Cheers

P.S. Java script is in a script tag at the bottom

nnnnnn · Accepted Answer · 2013-02-11 21:14:31Z

3

You've used a.selectedIndex to get the selected item for all three elements. Change that to use b.selectedIndex and c.selectedIndex in the appropriate spots. Or you can just say a.value, b.value and c.value directly rather than going via the options collection.

Also the id attributes in your html don't match what you're using in your JS.

Fix those problems and it works, as you can see here: http://jsfiddle.net/6WWdc/

P.S. It's a good idea to get in the habit of always passing the radix as the second parameter to parseInt(), like IntA = parseInt(A, 10); - otherwise if the input string has a leading 0 or leading 0x it might (depending on browser, and presense of a "use strict"; directive) be interpreted as octal or hexadecimal.

edited Feb 11, 2013 at 21:14

answered Feb 11, 2013 at 21:03

nnnnnn

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2 Comments

nzifnab Over a year ago

... Or you can extract common functionality and stop repeating yourself!

user2057440 Over a year ago

cheers i noticed the names after i put it on lol i was just using it as a example will try the code now cheers

wxactly · Accepted Answer · 2013-02-11 21:05:02Z

0

function myFunction()
{
    var a = document.getElementById("list1");
    var A = a.options[a.selectedIndex].value;

    var b = document.getElementById("list2");
    var B = b.options[b.selectedIndex].value; //changed to var b

    var c = document.getElementById("list3");
    var C = c.options[c.selectedIndex].value; //changed to var c

    //...
}

answered Feb 11, 2013 at 21:05

wxactly

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