获取对jQuery UI Widget所有实例的引用? [英] Get reference to all instances of jquery ui widget?

问题描述

我正在编写一个jQuery UI小部件，它仅包装了bootstrap popover插件，在该小部件中，您可以传递选项"singular"， 如果传入此参数，则应调用该插件的所有其他实例的函数.

I am writing a jquery UI widget that simply wraps the bootstrap popover plugin, In the widget you can pass in the option 'singular', if this is passed in then it should call a function of all other instances of the plugin.

类似

$('#one').myWidget();
$('#two').myWidget();
$('#three').myWidget();
$('#four').myWidget();

$('#one').myWidget('show'); //stuff from widget one is now visible
$('#two').myWidget('show'); //stuff from widget one and two are now visible
$('#three').myWidget('show'); //stuff from widget one, two and three are now visible
$('#two').myWidget('hide'); //stuff from widget one and three are now visible
$('#four').myWidget('show', {singular:true}); //stuff from widget four is now visible

所以，我想象显示函数看起来像:

So, I imagine the show function looking like:

show: function(options){
    options = options || {};

    if(options.singular){
        var instances = '????'; // how do I get all instances?
        $.each(instances, function(i, o){
            o.myWidget('hide');
        });
    }

    this.element.popover('show');

}

那么，问题是，我如何获得所有在其上具有myWidget小部件的元素的引用?

So, question being, how would I get a reference to all elements that have the myWidget widget on them?

推荐答案

您可以使用$(':ui-myWidget')，其中ui是窗口小部件的名称空间.它比使用诸如$('.ui-myWidget')之类的类选择器要慢，因此在创建窗口小部件时添加类仍然是一种好习惯.

You can use $(':ui-myWidget') where ui is your widget's namespace. It is slower than using a class selector like $('.ui-myWidget') so it is still good practice to add the class when your widget gets created.

jQuery UI对所有小部件都执行此操作，因此您可以通过$(':ui-progressbar')或$('.ui-progressbar')来获取每个进度条.

jQuery UI does this for all there widgets so you could get each progressbar by either $(':ui-progressbar') or $('.ui-progressbar').

此博客文章深入解释了

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