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Home/ Questions/Q 6229517
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T09:34:50+00:00 2026-05-24T09:34:50+00:00

I don’t believe any similar existing questions answer this one. I am developing a

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I don’t believe any similar existing questions answer this one.

I am developing a plugin that will turn <input type='checkbox' /> into a <div> with two toggle states. The basic idea for use is:

$('div .checkboxContainer').prettyBox();

The pseudo code for the plugin itself is:

$.fn.prettyBox = function(){
    return this.each(function(){
        $(this).find(':checkbox').each(function(){
           .. grab all event handlers on the current <input>
           .. create a new <div>
           .. attach all of the <input>'s event handlers to the <div>
           .. hide the <input>
           .. place the <div> where the <input> used to live
        });  
    };
};

Others who have asked similar questions have been concerned with copying single events, such as a click handler. To maintain plugin flexibility, I think it’s important that my code loop through everything attached to the input and copy it over.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-24T09:34:53+00:00Added an answer on May 24, 2026 at 9:34 am

    Try this

    $.fn.prettyBox = function(){
        return this.each(function(){
            $(this).find(':checkbox').each(function(){
               //Grabs all the events
               var events = $(this).data("events");
               var $div = $("<div />");
    
               //Loop through all the events
               $.each(events, function(i, event) {
                 //Loop through all the handlers attached for a event
                 $.each(event, function(j, h) {
                   //Bind the handler with the event 
                   $div.bind(i, h.handler);
                 });
               });
    
               $(this).hide().after($div);
            });  
        };
    };
    
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