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Editorial Team
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Asked: 2026-05-27T23:59:28+00:00

I have the following code below. Id like to replace the highlighted section with

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I have the following code below. Id like to replace the highlighted section with a single line of code that will run all that code. The reason behind this is because i reuse this code a lot, and would save me a lot of time by being able to copy and paste a single line of code.

Any ideas on the best way to write this?

This is the part i would like replacing with a single line of code.

$(".menu-item-1").show();
$(".menu-item-2").show();
$(".menu-item-3").show();
$(".menu-item-4").show();
$(".menu-item-5").show();
$(".menu-item-6").show();

And the entirety of the code below.

if (index == 2) {
$("#prev").show();
$("#next").show();
$(".menu-item-1").show();
$(".menu-item-2").show();
$(".menu-item-3").show();
$(".menu-item-4").show();
$(".menu-item-5").show();
$(".menu-item-6").show();
$("#p3-1").delay(1000).fadeIn("slow", function ()   {
    $("#p3-2").delay(500).fadeIn("slow", function ()    {
        $("#p3-3").delay(500).fadeIn("slow", function ()    {
            $("#p3-4").delay(500).fadeIn("slow", function ()    {
                $("#p3-5").delay(500).fadeIn("slow", function ()    {
                    $("#p3-6").delay(500).fadeIn("slow", function ()    {
                        $("#p3-7").delay(500).fadeIn("slow")
                        $("#next").show();
                    });
                });
            });
        });
    });
});
}
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1 Answer

  1. Editorial Team
    $(".menu-item-1").show();
$(".menu-item-2").show();
$(".menu-item-3").show();
$(".menu-item-4").show();
$(".menu-item-5").show();
$(".menu-item-6").show();

    should be 

    $(".menu-item-1, .menu-item-2, .menu-item-3, .menu-item-4, .menu-item-5, .menu-item-6").show();

    jQuery allows you to apply same effect to multiple selectors.

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