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Home/ Questions/Q 7078677
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T06:34:38+00:00 2026-05-28T06:34:38+00:00

I have the loop below: var myArray = []; $(this).children(‘a’).each(function () { myArray.push($(this).attr(‘href’)); });

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I have the loop below:

        var myArray = [];
        $(this).children('a').each(function () {
            myArray.push($(this).attr('href'));
        });

which fires 3 times. When I look inside that array, I see that there is only one (last added) item. Why ?

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  1. Editorial Team
    Editorial Team
    2026-05-28T06:34:38+00:00Added an answer on May 28, 2026 at 6:34 am

    Because you use declare myArray as a local variable. If you want the array’s values to persist, move var myArray = []; outside the common function.

    var myArray = []; // This variable is shared by all instances of somefunction
    $('#example').click(function() {
        $(this).children('a').each(function () {
            myArray.push($(this).attr('href'));  //myArray in the parent scope
        });
    });
    
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