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Home/ Questions/Q 8474629
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T17:40:51+00:00 2026-06-10T17:40:51+00:00

I’m building a jQuery plugin. Calling the plugin $(‘#box’).jQueryPlugin({user:’user123′}); JQUERY PLUGIN (function($){ $.fn.jQueryPlugin= function(options)

  • 0

I’m building a jQuery plugin.

Calling the plugin

$('#box').jQueryPlugin({user:'user123'});

JQUERY PLUGIN

(function($){  
    $.fn.jQueryPlugin= function(options) {  

        var  
          defaults = {  
            user: ''
          }

            var options = $.extend(defaults, options);
            var o = options; 

             $.ajax({
               type: "get",
               url: "http://api.domain.com/user/"+o.user,
               data: "",
               dataType: "jsonp",
               success: function(data){
                    var p = data;
                    console.log(p.location);
                    $(this).html(p.location);
               }
            });

          // returns the jQuery object to allow for chainability.  
          return this;  
    }  
})(jQuery);  

If I were to use the above, the console.log would show an error that it couldn’t write the p.location inside the div with id=”box”

How would I be able to get it so that it can write to whichever div is specified when calling the plugin?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T17:40:52+00:00Added an answer on June 10, 2026 at 5:40 pm

    this won’t have the context you expect in the success callback, so you just need to assign your div to a var so you can use it later..

    (function($){  
        $.fn.jQueryPlugin= function(options) {  
    
            var  
              defaults = {  
                user: ''
              }
    
                var options = $.extend(defaults, options);
                var o = options; 
    
                var $div = $(this);
    
                 $.ajax({
                   type: "get",
                   url: "http://api.domain.com/user/"+o.user,
                   data: "",
                   dataType: "jsonp",
                   success: function(data){
                        var p = data;
                        console.log(p.location);
                        $div.html(p.location); // now we have the original div;
                   }
                });
    
              // returns the jQuery object to allow for chainability.  
              return this;  
        }  
    })(jQuery);
    

    Another method would be to set the context option in the $.ajax call, see the context option in the jQuery docs.

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