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jQuery .wrap() not wrapping around a cloned element

Asked 2011-12-16T09:12:08.063
13

(function($) {
  $.extend({
    notify: function(options, duration) {
      var defaults = {
        inline: true,
        href: '',
        html: ''
      };
      var options = $.extend(defaults, options);

      var body = $('body'),
        container = $('<ul></ul>').attr('id', 'notification_area'),
        wrapper = '<li class="notification"></li>',
        clone;

      if (!body.hasClass('notifications_active')) {
        body.append(container).addClass('notifications_active');
      }

      if (options.inline == true && options.href) {
        clone = $(options.href).clone().wrap(wrapper);
      }

      clone.css('visibility', 'hidden').appendTo(container);

      var clone_height = 0 - parseInt(clone.outerHeight());
      clone.css('marginBottom', clone_height);

      clone.animate({
        marginBottom: 0
      }, 'fast', function() {
        clone.hide().css('visibility', 'visible').fadeIn('fast');
      });
    }
  });
})(jQuery);

$(function() {
  $('a').click(function() {
    $.notify({
      inline: true,
      href: '#alert'
    }, 3000)
  })
})
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

http://jsfiddle.net/sambenson/RmkEN/

In the above example I'm cloning an element and attempting to wrap it with and <li></li> but the clone isn't being wrapped at all. Why?

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1 Answer

23

The confusing part is that .wrap() returns the inner element, not the parent element.

So you have to use the parent object of the wrapped one as follows:

var $divA= $("<div/>").addClass('classA'),
    $divB= $("<div/>").addClass('classB');

console.log( $divA.wrap($divB).parent() );

($divA.parent() is equal to $divB after the wrapping)

So the key part is that $divA.wrap($divB) returns $divA, NOT $divB

see the reference:

This method returns the original set of elements for chaining purposes.

Please note: The elements DON'T have to be in the DOM, jQuery can operate on them without them already having been inserted into the DOM.

answered 2012-12-29T17:13:52.443

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