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How to move an element into another element?

I would like to move one DIV element inside another. For example, I want to move this (including all children):

<div id="source">
...
</div>

into this:

<div id="destination">
...
</div>

so that I have this:

<div id="destination">
  <div id="source">
    ...
  </div>
</div>
Just use $(target_element).append(to_be_inserted_element);
Or: destination.appendChild(source) using plain javascript
can we achieve this using CSS ? is that possible ?
@RajKumarSamala CSS can't alter the structure of the HTML, only its presentation.

C
Clonkex

You may want to use the appendTo function (which adds to the end of the element):

$("#source").appendTo("#destination");

Alternatively you could use the prependTo function (which adds to the beginning of the element):

$("#source").prependTo("#destination");

Example:

$("#appendTo").click(function() { $("#moveMeIntoMain").appendTo($("#main")); }); $("#prependTo").click(function() { $("#moveMeIntoMain").prependTo($("#main")); }); #main { border: 2px solid blue; min-height: 100px; } .moveMeIntoMain { border: 1px solid red; }

main
move me to main


A warning that this may not work correctly in jQuery mobile, as it may create another copy of the element instead.
does the appenTo create a copy or actually moves the whole div to the destination? (because if it copies, it would create erros when calling the div by the id)
Here is an excellent article on Removing, Replacing and Moving Elements in jQuery: elated.com/articles/jquery-removing-replacing-moving-elements
Note the jQuery documentation for appendTo states the element is moved: it will be moved into the target (not cloned) and a new set consisting of the inserted element is returned - api.jquery.com/appendto
You are not moving it, just appending. Below answer does a proper MOVE.
R
René Sackers

my solution:

MOVE:

jQuery("#NodesToMove").detach().appendTo('#DestinationContainerNode')

COPY:

jQuery("#NodesToMove").appendTo('#DestinationContainerNode')

Note the usage of .detach(). When copying, be careful that you are not duplicating IDs.


Best answer. Accepted answer creates a copy, doesn't move the element like the question asks for.
Sorry, but Andrew Hare's accepted answer is correct - the detach is unnecessary. Try it in Pixic's JSFiddle above - if you remove the detach calls it works exactly the same, i.e. it does a move, NOT a copy. Here's the fork with just that one change: jsfiddle.net/D46y5 As documented in the API: api.jquery.com/appendTo : "If an element selected this way is inserted into a single location elsewhere in the DOM, it will be moved into the target (not cloned) and a new set consisting of the inserted element is returned"
t
tyler.frankenstein

What about a JavaScript solution?

// Declare a fragment:
var fragment = document.createDocumentFragment();

// Append desired element to the fragment:
fragment.appendChild(document.getElementById('source'));

// Append fragment to desired element:
document.getElementById('destination').appendChild(fragment);

Check it out.


Why using createDocumentFragment instead of simply document.getElementById('destination').appendChild(document.getElementById('source'))?
C
Cᴏʀʏ

I just used:

$('#source').prependTo('#destination');

Which I grabbed from here.


Well, detach is useful when you want to hold on to the element and reinsert it later on, but in your example you reinsert it instantly anyway.
B
Bekim Bacaj

Ever tried plain JavaScript... destination.appendChild(source); ?

onclick = function(){ destination.appendChild(source); } div{ margin: .1em; } #destination{ border: solid 1px red; } #source {border: solid 1px gray; }

###
***


7
7 revs, 2 users 66%

If the div where you want to put your element has content inside, and you want the element to show after the main content:

  $("#destination").append($("#source"));

If the div where you want to put your element has content inside, and you want to show the element before the main content:

$("#destination").prepend($("#source"));

If the div where you want to put your element is empty, or you want to replace it entirely:

$("#element").html('<div id="source">...</div>');

If you want to duplicate an element before any of the above:

$("#destination").append($("#source").clone());
// etc.

A
AlexC

You can use:

To Insert After,

jQuery("#source").insertAfter("#destination");

To Insert inside another element,

jQuery("#source").appendTo("#destination");

S
Subodh Ghulaxe

You can use following code to move source to destination

 jQuery("#source")
       .detach()
       .appendTo('#destination');

try working codepen

function move() { jQuery("#source") .detach() .appendTo('#destination'); } #source{ background-color:red; color: #ffffff; display:inline-block; padding:35px; } #destination{ background-color:blue; color: #ffffff; display:inline-block; padding:50px; }

I am source
I am destination


D
Dan

If you want a quick demo and more details about how you move elements, try this link:

http://html-tuts.com/move-div-in-another-div-with-jquery

Here is a short example:

To move ABOVE an element:

$('.whatToMove').insertBefore('.whereToMove');

To move AFTER an element:

$('.whatToMove').insertAfter('.whereToMove');

To move inside an element, ABOVE ALL elements inside that container:

$('.whatToMove').prependTo('.whereToMove');

To move inside an element, AFTER ALL elements inside that container:

$('.whatToMove').appendTo('.whereToMove');

H
HMR

Old question but got here because I need to move content from one container to another including all the event listeners.

jQuery doesn't have a way to do it but standard DOM function appendChild does.

//assuming only one .source and one .target
$('.source').on('click',function(){console.log('I am clicked');});
$('.target')[0].appendChild($('.source')[0]);

Using appendChild removes the .source and places it into target including it's event listeners: https://developer.mozilla.org/en-US/docs/Web/API/Node.appendChild


f
frb

You may also try:

$("#destination").html($("#source"))

But this will completely overwrite anything you have in #destination.


A
Alireza

You can use pure JavaScript, using appendChild() method...

The appendChild() method appends a node as the last child of a node. Tip: If you want to create a new paragraph, with text, remember to create the text as a Text node which you append to the paragraph, then append the paragraph to the document. You can also use this method to move an element from one element to another. Tip: Use the insertBefore() method to insert a new child node before a specified, existing, child node.

So you can do that to do the job, this is what I created for you, using appendChild(), run and see how it works for your case:

function appendIt() { var source = document.getElementById("source"); document.getElementById("destination").appendChild(source); } #source { color: white; background: green; padding: 4px 8px; } #destination { color: white; background: red; padding: 4px 8px; } button { margin-top: 20px; }

Source

Destination


A
AI Perera

I noticed huge memory leak & performance difference between insertAfter & after or insertBefore & before .. If you have tons of DOM elements, or you need to use after() or before() inside a MouseMove event, the browser memory will probably increase and next operations will run really slow.

The solution I've just experienced is to use inserBefore instead before() and insertAfter instead after().


K
Kamil Kiełczewski

dirty size improvement of Bekim Bacaj answer

div { border: 1px solid ; margin: 5px }

click me
...


R
RayLuo

For the sake of completeness, there is another approach wrap() or wrapAll() mentioned in this article. So the OP's question could possibly be solved by this (that is, assuming the <div id="destination" /> does not yet exist, the following approach will create such a wrapper from scratch - the OP was not clear about whether the wrapper already exists or not):

$("#source").wrap('<div id="destination" />')
// or
$(".source").wrapAll('<div id="destination" />')

It sounds promising. However, when I was trying to do $("[id^=row]").wrapAll("<fieldset></fieldset>") on multiple nested structure like this:

<div id="row1">
    <label>Name</label>
    <input ...>
</div>

It correctly wraps those <div>...</div> and <input>...</input> BUT SOMEHOW LEAVES OUT the <label>...</label>. So I ended up use the explicit $("row1").append("#a_predefined_fieldset") instead. So, YMMV.