1926

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>
4
  • 27
    Just use $(target_element).append(to_be_inserted_element); Commented Jun 27, 2013 at 20:07
  • 5
    Or: destination.appendChild(source) using plain javascript
    – luke
    Commented Nov 15, 2016 at 22:10
  • can we achieve this using CSS ? is that possible ? Commented Mar 8, 2017 at 15:00
  • 11
    @RajKumarSamala CSS can't alter the structure of the HTML, only its presentation. Commented Mar 8, 2017 at 18:49

16 Answers 16

1948

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;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="main">main</div>
<div id="moveMeIntoMain" class="moveMeIntoMain">move me to main</div>

<button id="appendTo">appendTo main</button>
<button id="prependTo">prependTo main</button>

6
  • 25
    A warning that this may not work correctly in jQuery mobile, as it may create another copy of the element instead. Commented Jun 19, 2012 at 14:07
  • 33
    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)
    – user1031721
    Commented Jul 12, 2012 at 22:04
  • 55
    Here is an excellent article on Removing, Replacing and Moving Elements in jQuery: elated.com/articles/jquery-removing-replacing-moving-elements
    – xhh
    Commented Dec 3, 2012 at 7:55
  • 47
    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
    – John K
    Commented Jan 13, 2014 at 1:10
  • 18
    You are not moving it, just appending. Below answer does a proper MOVE.
    – Aaron
    Commented May 28, 2014 at 3:42
1028

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.

2
  • 94
    Best answer. Accepted answer creates a copy, doesn't move the element like the question asks for.
    – paulscode
    Commented Dec 17, 2013 at 19:02
  • 109
    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" Commented Feb 4, 2014 at 4:34
163

Use a vanilla 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.

2
  • 30
    Why using createDocumentFragment instead of simply document.getElementById('destination').appendChild(document.getElementById('source'))? Commented Feb 16, 2016 at 1:45
  • Since the document fragment is in memory and not part of the main DOM tree, appending children to it does not cause page reflow (computation of element's position and geometry). Historically, using document fragments could result in better performance. Source - developer.mozilla.org/en-US/docs/Web/API/Document/… Commented Oct 31, 2023 at 11:45
118

Try plain JavaScript? destination.appendChild(source);.

addEventListener("click", () => document.getElementById("destination").appendChild(document.getElementById("source")));
div {
  margin: .1em;
}

#destination {
  border: solid 1px red;
}

#source {
  border: solid 1px gray;
}
<div id="destination">
  ###
</div>
<div id="source">
  ***
</div>

1
  • 2
    “Editing is important for keeping posts clear, relevant, and up-to-date. If you are not comfortable with the idea of your contributions being collaboratively edited by other trusted users, this may not be the site for you.” — from the help center. Commented May 21, 2023 at 3:08
111

I just used:

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

Which I grabbed from here.

1
  • 10
    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.
    – Tim Büthe
    Commented Sep 27, 2012 at 15:56
100

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.
0
43

You can use:

To insert after,

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

To insert inside another element,

jQuery("#source").appendTo("#destination");
27

You can use the following code to move the source to the destination:

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

Try the 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;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="source">
I am source
</div>

<div id="destination">
I am destination
</div>

<button onclick="move();">Move</button>

24

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');
15

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 the 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 its event listeners: Node.appendChild() (MDN)

10

You may also try:

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

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

0
10

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;
}
<div id="source">
  <p>Source</p>
</div>

<div id="destination">
  <p>Destination</p>
</div>

<button onclick="appendIt()">Move Element</button>

7

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().

1
  • On what browser, incl. version, and operating system, etc. was this observed? Commented Nov 12, 2022 at 4:51
6

Dirty size improvement of Bekim Bacaj's answer:

div { border: 1px solid ; margin: 5px }
<div id="source" onclick="destination.appendChild(this)">click me</div>
<div id="destination" >...</div>

3

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.

0

The .appendChild does precisely that - basically a cut& paste.

It moves the selected element and all of its child nodes.

Not the answer you're looking for? Browse other questions tagged or ask your own question.