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Home/ Questions/Q 562759
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Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T12:35:19+00:00 2026-05-13T12:35:19+00:00

So that it can be show or hide this way: $(selector).pluginName(‘show’) $(selector).pluginName(‘hide’) The problem

  • 0

So that it can be show or hide this way:

$(selector).pluginName('show')
$(selector).pluginName('hide')

The problem is how do I know which one to show or hide?

I’m now doing it this way:

$.fn.pluginName = function(opts) {  
    var conf = $.extend({},opts);
    return this.each(function() { 
        if(conf && 'conf' == conf.type)
        {
            //ClassName is defined elsewhere
            new ClassName(conf,$(this));
        }
        else
        {
            //**show or hide corresponding instance**
        }
    });
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T12:35:19+00:00Added an answer on May 13, 2026 at 12:35 pm

    You can use data to associate your object with the DOM element it belongs to:

    $.fn.pluginName = function(opts) {
      if(typeof(opts) == "string"){
        this.each(function(){
          // Get the associated obj
          var obj = $(this).data('ClassName');
          if(obj){
            if(opts == "show") obj.myShowMethod();
            else if (opts == "hide") obj.myHideMethod();
          }
        })
      } else {
        var conf = $.extend({},opts);  
        this.each(function() { 
          var obj = new ClassName(conf,$(this));
          // Associate your object with this DOM element
          $(this).data('ClassName', obj);
        });
      }
      return this; // Don't break the chain
    }
    

    Also check out, Starter for jQuery, which uses the data pattern to associate the object with the DOM element.

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