jQuery .wrap() not wrapping around a cloned elemen

2019-03-18 16:34发布

(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)
    })
})

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.

Any ideas why?

2条回答
▲ chillily
2楼-- · 2019-03-18 17:23

The confusing part might have been to You 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 already having been inserted into the DOM.

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仙女界的扛把子
3楼-- · 2019-03-18 17:30

The key is this line in the .wrap() documentation:

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

.wrap() only operates on an element already in the DOM. So, you will need to insert it, then wrap it.

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