从jQuery选择中删除没有id的元素？

Question

如何id从调用插件的集合(而不是从DOM本身)中删除没有属性的元素？

<span id="id737293" class="attach"></span>
<div class="attach"></span>

jQuery调用和插件:

$('.attach').attach();

(function($) {

    $.fn.attach = function(options) {
       // Remove elements (without id) on which the plugin was invoked
       var valid = ???;

       // Loop each valid element (with id attribute) and process
       valid.each(function() {
       });

       return this; // Maintain chainability
    };

})(jQuery);

Answer 1

使用.filter删除没有ID的元素.

(function($) {

    $.fn.attach = function(options) {
       // Remove elements (without id) on which the plugin was invoked
       var valid = this.filter(function() { return !!this.id; });

       // Loop each valid element (with id attribute) and process
       valid.each(function() {               
       });

       return this; // Maintain chainability
    };

})(jQuery);

$('.attach').attach();

http://jsfiddle.net/5Kn9W/2/

或var valid = this.not(':not([id])');- http://jsfiddle.net/5Kn9W/1/