JavaScript moving element in the DOM

Question

Let's say I have three <div> elements on a page. How can I swap positions of the first and third <div>? jQuery is fine.

Answer 1

There's no need to use a library for such a trivial task:

var divs = document.getElementsByTagName("div");   // order: first, second, third
divs[2].parentNode.insertBefore(divs[2], divs[0]); // order: third, first, second
divs[2].parentNode.insertBefore(divs[2], divs[1]); // order: third, second, first

This takes account of the fact that getElementsByTagName returns a live NodeList that is automatically updated to reflect the order of the elements in the DOM as they are manipulated.

You could also use:

var divs = document.getElementsByTagName("div");   // order: first, second, third
divs[0].parentNode.appendChild(divs[0]);           // order: second, third, first
divs[1].parentNode.insertBefore(divs[0], divs[1]); // order: third, second, first

and there are various other possible permutations, if you feel like experimenting:

divs[0].parentNode.appendChild(divs[0].parentNode.replaceChild(divs[2], divs[0]));

for example :-)

Answer 2

Trivial with jQuery

$('#div1').insertAfter('#div3');
$('#div3').insertBefore('#div2');

If you want to do it repeatedly, you'll need to use different selectors since the divs will retain their ids as they are moved around.

$(function() {
    setInterval( function() {
        $('div:first').insertAfter($('div').eq(2));
        $('div').eq(1).insertBefore('div:first');
    }, 3000 );
});

Answer 3

.before and .after

Use modern vanilla JS! Way better/cleaner than previously. No need to reference a parent.

const div1 = document.getElementById("div1");
const div2 = document.getElementById("div2");
const div3 = document.getElementById("div3");

div2.after(div1);
div2.before(div3);

All modern browsers are supported!

Browser Support

Answer 4

jQuery.fn.swap = function(b){ 
    b = jQuery(b)[0]; 
    var a = this[0]; 
    var t = a.parentNode.insertBefore(document.createTextNode(''), a); 
    b.parentNode.insertBefore(a, b); 
    t.parentNode.insertBefore(b, t); 
    t.parentNode.removeChild(t); 
    return this; 
};

and use it like this:

$('#div1').swap('#div2');

if you don't want to use jQuery you could easily adapt the function.

Answer 5

var swap = function () {
    var divs = document.getElementsByTagName('div');
    var div1 = divs[0];
    var div2 = divs[1];
    var div3 = divs[2];

    div3.parentNode.insertBefore(div1, div3);
    div1.parentNode.insertBefore(div3, div2);
};

This function may seem strange, but it heavily relies on standards in order to function properly. In fact, it may seem to function better than the jQuery version that tvanfosson posted which seems to do the swap only twice.

What standards peculiarities does it rely on?

insertBefore Inserts the node newChild before the existing child node refChild. If refChild is null, insert newChild at the end of the list of children. If newChild is a DocumentFragment object, all of its children are inserted, in the same order, before refChild. If the newChild is already in the tree, it is first removed.

Answer 6

Jquery approach mentioned on the top will work. You can also use JQuery and CSS .Say for e.g on Div one you have applied class1 and div2 you have applied class class2 (say for e.g each class of css provides specific position on the browser), now you can interchange the classes use jquery or javascript (that will change the position)

Answer 7

Sorry for bumping this thread I stumbled over the "swap DOM-elements" problem and played around a bit

The result is a jQuery-native "solution" which seems to be really pretty (unfortunately i don't know whats happening at the jQuery internals when doing this)

The Code:

$('#element1').insertAfter($('#element2'));

The jQuery documentation says that insertAfter() moves the element and doesn't clone it