IE getElementById uses name not id??

Hey guys,

I'm having some troubles understanding something, and wondering if it's a bug in IE or what (I don't have netscape at work...).

I have a multipage form that is written in PHP/Javascript, and the Javascript takes advantage of DOM. As far as the PHP is concerned, it is merely there to retain values from page to page using fieldforwarder.php (http://www.zend.com/zend/spotlight/code-gallery-wade7.php#Heading9).

So what fieldforwarder does is creates hidden input fields of the old values from the different pages.

But then I have my actual fields which are not hidden on the same page. So essentially I have two fields with the same name. So on the 'active' field, I assign an id to it so that I can reference it using getElementById().

However, I'm finding that getElementById() will return the hidden field, which is not what I want!

I know this because first of all I'm getting the wrong values from my boxes. Secondly, when I get the element, I alert element.getAttribute("id") and it comes up blank (because the hidden fields don't have an id).

So.... help?

Thanks,
Sadiq.

This might be worth experimenting with: uniqueID Property (http://msdn.microsoft.com/workshop/author/dhtml/reference/properties/uniqueid.asp)

IE does not make a diffenrence between name and id attribute in getElementById, the 1st element with either attribute set to the searched value will be the element returned.

Hmm, doesn't ie actually return a collection of all elements with that id when used multiple times on page?

Yeah it does; stupid thing. Any freshman psyche student could tell that "id" and "name" are not the same - your name is part of your self-identify, which is a facet of "ego" not "id" :p

Anyway .. the solution is don't use the same value for the ID of one element and the NAME of another.

Anyway .. the solution is don't use the same value for the ID of one element and the NAME of another.
Yep that's what I'm doing now.. what a piss off though..

Thanks everyone for your help.

Sadiq.