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Home/ Questions/Q 6650419
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T00:53:27+00:00 2026-05-26T00:53:27+00:00

The following function should reveal a panel when the input has 3 or more

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The following function should reveal a panel when the input has 3 or more characters inside it and is not equal to the placeholder.

$('#search').keydown(function(){
            $('#header .suggestion').removeClass('visible');
            var slength = $(this).val();
            if(slength.length > 2 && slength != $(this).attr('placeholder')){
                //activate AJAX script
                $('#header .results').addClass('visible');
            }
            else {
                $('#header .results').removeClass('visible');
            }

    });

Instead it doesn’t appear until 5 characters are displayed. Why.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T00:53:27+00:00Added an answer on May 26, 2026 at 12:53 am

    The keydown and keypress events are called BEFORE the browser has processed the input. During these events, the developer can add a method to prevent the user from entering text.

    So, when you’re listening to the keydown function, the $(this).val() does NOT return the “expected” value.

    An image (the user has already focused on the input field):

    { } <-- Empty input field
        [X] <-- Use presses the X key
        **KEYDOWN**
        **KEYPRESS**
    {X} <-- "x" appears in the text field
        **KEYUP**
    

    To get the “expected” value of the input, you have to use the keyup event instead of keydown.

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