How to move an element into another element

Asked 2023-09-21 08:08:18 View 305,683

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); - anyone
  • Or: destination.appendChild(source) using plain javascript - anyone
  • can we achieve this using CSS ? is that possible ? - anyone
  • @RajKumarSamala CSS can't alter the structure of the HTML, only its presentation. - anyone

Answers

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>

Answered   2023-09-21 08:08:18

  • A warning that this may not work correctly in jQuery mobile, as it may create another copy of the element instead. - anyone
  • 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) - anyone
  • Here is an excellent article on Removing, Replacing and Moving Elements in jQuery: elated.com/articles/jquery-removing-replacing-moving-elements - anyone
  • 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 - anyone
  • You are not moving it, just appending. Below answer does a proper MOVE. - anyone

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.

Answered   2023-09-21 08:08:18

  • Best answer. Accepted answer creates a copy, doesn't move the element like the question asks for. - anyone
  • 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" - anyone

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.

Answered   2023-09-21 08:08:18

  • Why using createDocumentFragment instead of simply document.getElementById('destination').appendChild(document.getElementById('source'))? - anyone

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>

Answered   2023-09-21 08:08:18

  • “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. - anyone

I just used:

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

Which I grabbed from here.

Answered   2023-09-21 08:08:18

  • 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. - anyone

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.

Answered   2023-09-21 08:08:18

You can use:

To insert after,

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

To insert inside another element,

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

Answered   2023-09-21 08:08:18

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>

Answered   2023-09-21 08:08:18

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');

Answered   2023-09-21 08:08:18

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)

Answered   2023-09-21 08:08:18

You may also try:

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

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

Answered   2023-09-21 08:08:18

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>

Answered   2023-09-21 08:08:18

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().

Answered   2023-09-21 08:08:18

  • On what browser, incl. version, and operating system, etc. was this observed? - anyone

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>

Answered   2023-09-21 08:08:18

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.

Answered   2023-09-21 08:08:18

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

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

Answered   2023-09-21 08:08:18