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
15 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>
6
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)
– user1031721Here is an excellent article on Removing, Replacing and Moving Elements in jQuery: elated.com/articles/jquery-removing-replacing-moving-elements
– xhhNote 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 KYou are not moving it, just appending. Below answer does a proper MOVE.
– Aaron
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
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”
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);
1
Why using createDocumentFragment instead of simply document.getElementById(‘destination’).appendChild(document.getElementById(‘source’))?
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; }
<!DOCTYPE html>
<html>
<body>
<div id="destination">
###
</div>
<div id="source">
***
</div>
</body>
</html>
0
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
You can use:
To Insert After,
jQuery("#source").insertAfter("#destination");
To Insert inside another element,
jQuery("#source").appendTo("#destination");
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');
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;
}
<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>
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
You may also try:
$("#destination").html($("#source"))
But this will completely overwrite anything you have in #destination
.
0
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()
.
0
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>
dirty size improvement of Bekim Bacaj answer
div { border: 1px solid ; margin: 5px }
<div id="source" onclick="destination.appendChild(this)">click me</div>
<div id="destination" >...</div>
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.
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.