jQuery virtual id?

Question

Is there something in jQuery to uniquely identify a DOM node, if the HTML markup does not provide an id, i.e. if $(NODE).attr("id") returns undefined?

Goal: exclude a specific node of a specific class.

jQuery(document).ready(function($){
$(".accordion-right-content").hide();
    $(".article-image img").click(function() {
    $(".accordion-right-content").each(function(i,v) {
        // the following comparison does not work because of the missing ID
        if ($(v).attr("id") !== $(this).parent().parent().children(".accordion-right-content").attr("id")) {
            $(v).hide();
        } else {
            $(v).show();
        }
    });
});

Except the specific $(this).parent().parent().children(".accordion-right-content") to be hidden before shown again.

Can you please write the html or create a fiddle to help us understand your problem? — Rupam Datta, Apr 15 '14 at 05:04
Each DOM node itself is unique. If you provide a more complete example and a better explanation, you will get more useful answers. — Felix Kling, Apr 15 '14 at 05:04

score 0 · Answer 1 · answered Apr 15 '14 at 05:27

0

You should use not operator like this:

if (!$(NODE).attr("id")){
//do stuff here
}

And the following, you're having wrong:

if ($(v).attr("id") !== $(this).parent().parent().children(".accordion-right-content").attr("id")) {

Should be this:

if ($(this).attr("id") !== $(this).parent().parent().children(".accordion-right-content").attr("id")) {

answered Apr 15 '14 at 05:27

Bhojendra Rauniyar

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You should be using `prop` instead of `attr`. – Jeremy J Starcher Apr 15 '14 at 05:56
why prop? attr is okay. – Bhojendra Rauniyar Apr 15 '14 at 06:01
http://stackoverflow.com/questions/5874652/prop-vs-attr There was a time when jQuery horribly, horribly muddled these two concepts. They are different, so they should be treated differently. – Jeremy J Starcher Apr 15 '14 at 06:02

jQuery virtual id?

1 Answers1