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Home/ Questions/Q 3307418
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T21:24:14+00:00 2026-05-17T21:24:14+00:00

Here’s my code: function hideColumnAndShowOther(columnToHide, columnToShow) { $(columnToHide).fadeTo(slow, 0.0, function() { $(columnToShow).fadeIn(slow); }); }

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Here’s my code:

function hideColumnAndShowOther(columnToHide, columnToShow) {
$(columnToHide).fadeTo("slow", 0.0, function() {
                       $(columnToShow).fadeIn("slow");
                       });
}

In this case the callback function isn’t called. I have used the firebug tool to get the root of the problem. In the callback function the ‘columnToShow’ variable is not present. I think it is logical because it is a separate block, but how can I solve this then? Do you have any tip?

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-17T21:24:15+00:00Added an answer on May 17, 2026 at 9:24 pm

    The problem is the columnToShow variable is out of scope. It means nothing at that point. Instead try $(this).fadeIn(“slow”).

    The $(this) refers to the element that you have just faded as the callback function is attached to that element.

    EDIT: (misread the question)

    If you change it something like the following it should work.

    function hideColumnAndShowOther(columnToHide, columnToShow) {
        var showColumn = columnToShow;
        $(columnToHide).fadeTo("slow", 0.0,
            function() {
               $(showColumn).fadeIn("slow");
            });
    }
    
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