Javascript code works, but "not defined" warning in Firebug

Question

I have javascript code below, it's working but within Firebug it says

document.form1.element[i] is not defined

and then it works fine

function set_action(){

for (var i=0;i<=3;i++)
{

    if (document.form1.PayType[i].checked == true)

        {
            var billerid = document.form1.billerid[i].value;
                            document.form1.action = billerid +"_mypag.htm";
        }
}

and my html markup is as below

<form name="form1" action="...">
<input name="PayType" type="radio" value="0" id="ultipay" class="radiobtn" checked/>
<select name="billerid" class="dropbox">
<option>item1</Option>...
</select>
<input name="PayType" type="radio" value="1" id="ultipay" class="radiobtn"/>
<select name="billerid" class="dropbox">
<option>item1</Option>
</select>
<input name="PayType" type="radio" value="2" id="ultipay" class="radiobtn"/>
<select name="billerid" class="dropbox">
<option>item1</Option>...
</select>
<input name="PayType" type="radio" value="3" id="ultipay" class="radiobtn"/>
<select name="billerid" class="dropbox">
<option>item1</Option>...
</select>
<input type="button" onclick="set_action()" value="submit">
</form>

I don't know why I am getting this error.

You only have 4 elements, but iterate 7 (or, even worse, 8 - thanks @Marcel) of them? — Bergi
– Bergi, Commented Jan 21, 2013 at 13:36
You should be using document.getElementById, not document.form.xxx. — jbabey
– jbabey, Commented Jan 21, 2013 at 13:42

Roland Mai · Accepted Answer · 2013-01-21 13:41:18Z

2

If you have only one radio button named PayType, then you need to address it with document.form1.PayType. It is addressed as an array document.form1.PayType[i] iff there are multiple radio buttons with the same name. For instance:

<input name="PayType" type="radio" value="0" id="ultipay0" class="radiobtn" checked="checked" />
<input name="PayType" type="radio" value="1" id="ultipay1" class="radiobtn" />

answered Jan 21, 2013 at 13:41

Roland Mai

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