i found that i can gram the id's like this:

var unic = $('div.hidde').find('li').map(function(i, v) { return this.id; }).get();
var unic1 = $('div.hi').find('li').map(function(i, v) { return this.id; }).get();

and i can bine them using this script:

var intersection = [];       
$(liList).each(function () {
    for (i in unic) {
        for (j in unic1) {
            if (unic[i] == unic1[j]) intersection.push(unic[i]);
        }
    }
});
alert(intersection);

another questions is how to remove the <li>'s with the id's returned by intersection ??

i found that i can gram the id's like this:

var unic = $('div.hidde').find('li').map(function(i, v) { return this.id; }).get();
var unic1 = $('div.hi').find('li').map(function(i, v) { return this.id; }).get();

and i can bine them using this script:

var intersection = [];       
$(liList).each(function () {
    for (i in unic) {
        for (j in unic1) {
            if (unic[i] == unic1[j]) intersection.push(unic[i]);
        }
    }
});
alert(intersection);

another questions is how to remove the <li>'s with the id's returned by intersection ??

• javascript
• jquery

• you might want to mention which JavaScript framework / library you're using here. Is it jQuery? – Bruce Commented Mar 18, 2011 at 6:46

2 Answers 2

If intersection is a list of IDs, it's really simple.

Using straight DOM methods (no jQuery), you can use getElementById to look up each element, parentNode to get its parent node, and then removeChild to remove it:

var index;
for (index = 0; index < intersection.length; ++index) {
    elm = document.getElementById(intersection[index]);
    elm.parentNode.removeChild(elm);
}

But it's worth noting that jQuery works around bugs in IE6's and IE7's implementation of getElementById, so if you're already using jQuery, you may want to use it for this as well to get the advantage of those workarounds.

Or use jQuery's remove:

var index;
for (index = 0; index < intersection.length; ++index) {
    $("#" + intersection[index]).remove();
}

Or if you like, use jQuery's each with remove:

$.each(intersection, function(index, value) {
    $("#" + value).remove();
});

Update 2011/04/29: Or you can take advantage of Array#join to create a selector in the form #id1, #id2, #id3 and them remove them with one call:

$('#' + intersection.join(', #')).remove();

Update: There's some discussion of making sure that the elements are li elements. If you really need to do that, here are the three above, modified:

Straight DOM (added the if):

var index;
for (index = 0; index < intersection.length; ++index) {
    elm = document.getElementById(intersection[index]);
    if (elm.tagName === "LI") {
        elm.parentNode.removeChild(elm);
    }
}

jQuery options (literally just add "li" in front of "#" to say "the element with this ID, but only if it's a li"):

For loop:

var index;
for (index = 0; index < intersection.length; ++index) {
    $("li#" + intersection[index]).remove();
}

Using each:

$.each(intersection, function(index, value) {
    $("li#" + value).remove();
});

Update 2011/04/29: Using Array#join:

$('li#' + intersection.join(', li#')).remove();

intersection has a list of all ids present in both arrays.

$('li').filter(function() {
    return $.inArray(this.id, intersection) !== -1;
}).remove();

jsFiddle.