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Home/ Questions/Q 6161375
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T21:28:05+00:00 2026-05-23T21:28:05+00:00

Why is it not using the Math.Random I put in there. Thanks in advance.

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Why is it not using the Math.Random I put in there. Thanks in advance.

    $(function () {
    $.ajax({
    type: "GET",
    url: "myFakeChannelData.xml",
    dataType: "xml",
    success: changeChannel
    });
});

function changeChannel(xml) {
    $('#layer').fadeOut(1000);
var $limit = 4;
    $(xml).find("Channel").each(function($limit) {  
    var $channel = $(this);
    var image = $channel.attr('image');
    $("#click").click(function () {
    $(".layer-container").empty();
        $(".layer-container").append('<div class="layer1">' + 
    '<img class="" alt=""   src="' + image + '" />' +  '</div></div>');
    $(".layer-container").append('<div class="layer2">' + '<img     class=""            alt=""  src="' + image + '" />' +  '</div></div>');
     $(".layer-container").append('<div class="layer3">' + '<img class="" alt=""    src="' + image + '" />' +  '</div></div>');
    $(".Channel").fadeIn(1000);
    var random = Math.floor(limit * Math.random());
    var myNewChannel = remix(random);
    myNewChannel(random);
    });
    });
}
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  1. Editorial Team
    Editorial Team
    2026-05-23T21:28:05+00:00Added an answer on May 23, 2026 at 9:28 pm

    You’re referring to “limit” in that expression, but it should be “$limit”.

     var random = Math.floor($limit * Math.random());
    

    It was probably always coming out zero, since the undefined “limit” would be cast to zero for the multiplication.

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