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Home/ Questions/Q 8763423
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T15:46:41+00:00 2026-06-13T15:46:41+00:00

I have this code: var mods1 = [‘a’, ‘b’, ‘c’, ‘d’]; var mods2 =

  • 0

I have this code:

var mods1 = ['a', 'b', 'c', 'd'];
var mods2 = ['1', '2', '3', '4'];
var mods3 = ['js', 'hates', 'me', ':('];


jQuery('div').append('<ul class="modlist"></ul>');
jQuery.each(mods1, function(i) {
    jQuery('<li/>').html(mods1[i]).appendTo('.modlist');
});

I need to use a different variable (mods1, modsN,…) depending of the URL where this script is used.

The URLs always have the same structure:

http://www.example.com/NUMBER/
http://www.example.com/NUMBER/example/
http://www.example.com/NUMBER/example/example

The “number” part is the one I can use to differentiate between pages.

jsFiddle with my example code running: http://jsfiddle.net/RZHxz/

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T15:46:42+00:00Added an answer on June 13, 2026 at 3:46 pm

    Try this:

    var url = "/1/example"; // location.pathname
    var id = url.split("/")[1];
    
    mods = [
        ['a', 'b', 'c', 'd'],
        ['1', '2', '3', '4'],
        ['js', 'hates', 'me', ':(']
    ];
    
    jQuery('div').append('<ul class="modlist"></ul>');
    jQuery.each(mods[id], function(i,v) {
        jQuery('<li/>').html(v).appendTo('.modlist');
    });
    

    I’ve made the mods into an array as that’s how you’re effectively accessing them anyway, also, the each() function passes the value to the handling function as a parameter, so you don’t need to go off to the array again.

    Example fiddle

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