Reputation: 12520
Fiddle here: http://jsfiddle.net/yv1vLhd9/
For whatever reasons, suppose I want to replace three dom elements with three others using JQuery, but I want to transfer one or more data attributes from the first elements onto their corresponding replacements.
Given the following HTML:
<div class='original' data-custom="dog"></div>
<div class='original' data-custom="cat"></div>
<div class='original' data-custom="sheep"></div>
And the following CSS:
div {
width: 100px;
height: 100px;
}
div:before {
content: attr(data-custom);
}
.original {
border: blue solid 1px;
}
.new {
border: pink solid 1px;
}
How can I make sure there is a new element with each of the custom data attributes?
For example:
$(document).ready( function(){
var $originalEl = $('.original')
var originalData = $originalEl.data('custom')
var replacement = '<div class="new" data-custom="' + originalData + '"></div>'
$originalEl.after(replacement).hide()
});
But this creates three new data-custom="dog" attributes.
Upvotes: 0
Views: 42
Reputation: 12508
jQuery provides a host of DOM manipulation methods that will help you achieve the desired input. As noted by some of the other answers, you'll need to traverse over the set of matched elements using $.each
. In addition, rather than creating new elements from strings, you can use the .clone()
method to:
$(function () {
$('.original').each(function (idx, elem) {
$(elem)
.clone()
.removeClass('original')
.addClass('new')
.insertAfter($(elem))
.prev()
.hide();
});
});
div {
width: 100px;
height: 100px;
}
div:before {
content: attr(data-custom);
}
.original {
border: blue solid 1px;
}
.new {
border: pink solid 1px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='original' data-custom="dog"></div>
<div class='original' data-custom="cat"></div>
<div class='original' data-custom="sheep"></div>
API Method References:
Upvotes: 1
Reputation: 3509
http://jsfiddle.net/VixedS/wz95hh3r/1/
$(document).ready(function(){
$('.original').each(function(){
$(this).hide().clone().attr('class','new').show().appendTo('body');
})
});
Upvotes: 1
Reputation: 362
You must use $.each
for the element.
Starting after you declare $originalEl
$originalEl.each(function(){
// code here
})
Here is your code updated http://jsfiddle.net/yv1vLhd9/4/
Upvotes: 4