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Home/ Questions/Q 780865
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T20:08:30+00:00 2026-05-14T20:08:30+00:00

I’ve been messing around with jQuery plugin code and I’m trying to put all

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I’ve been messing around with jQuery plugin code and I’m trying to put all of my common variables into a single object for easy access. I have included a few examples below on how I’ve done this, but I’m wondering how others deal with this problem.

Lets say I have this

var x = function(options){
 var defaults = {
  ulist   : $('ul#list'),
  listLen : $('ul#list').children().length
 }
 $.extend(options, defaults);
 // do other stuff
}

What I’m trying to do is use the ulist object in as a base, then find the number of li‘s

I guess I could do this:

var x = function(options){
 var defaults = {
  ulist   : $('ul#list'),
  listLen : 0
 }
 defaults.listLen = defaults.ulist.children().length;
 $.extend(options, defaults);
 // do other stuff
}

or this:

var x = function(options){
 var defaults = {
  ulist : $('ul#list')
 };
 var defaults2 = {
  listLen : defaults.ulist.children().length
 }
 $.extend(defaults, defaults2);
 $.extend(options, defaults);
 // do other stuff
}

The above code samples are just thrown together, and only meant to get the idea across to you. Anyway, is there a better way to do this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T20:08:31+00:00Added an answer on May 14, 2026 at 8:08 pm

    I would say this version that you posted, but no need to initialize listLen to 0.

    var x = function(options){
     var defaults = {
      ulist   : $('ul#list'),
      listLen : 0    // Remove this line
     }
     defaults.listLen = defaults.ulist.children().length;
     $.extend(options, defaults);
     // do other stuff
    }
    

    EDIT:

    Your second solution would work too, but no need to do two calls to $.extend(). It can accept more than one object.

    $.extend(options, defaults, defaults2);
    

    Your first solution still seems better.


    EDIT:

    (As pointed to, you can use a function. Although I’d still assign the function outside the initialization of the defaults object, so you can add the function call operator () at the end, and call it like a property.

    var x = function(options){
        var defaults = {
            ulist: $('ul#list')
        } 
    
        defaults.listLen = function() {return defaults.ulist.children().length}();
    }
    

    Now you can access defaults.listLen like a property and get the result of the function call, sort of like calling .length on a jQuery object.

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