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Home/ Questions/Q 7566169
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T14:20:37+00:00 2026-05-30T14:20:37+00:00

I have this jQuery script: $(.show-div).data(‘loaded’,false).click(function() { $.post(script.php, {id_1: var_id_1, id_2: var_id_2}, function(data){ $(this).data(‘loaded’,true);

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I have this jQuery script:

$(".show-div").data('loaded',false).click(function() {
    $.post("script.php", {id_1: var_id_1, id_2: var_id_2}, function(data){
        $(this).data('loaded',true);
        $("#display").html(data);
    });

    $("#display").toggle('fast');
    return false;
});

That loads a page (script.php) and displays its content inside the #display div from the current page. Now, what I’m trying to do is load script.php only the first time the .show-div is clicked, but instead it gets loaded each time. Why? What am I doing wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T14:20:39+00:00Added an answer on May 30, 2026 at 2:20 pm
    $(".show-div").click(function() {
        if (!$(this).data('loaded'))  // <====
        {
            $.post("script.php", {id_1: var_id_1, id_2: var_id_2}, function(data){
                $("#display").html(data);
                $(this).data('loaded',true);
            });
        }
    
        $("#display").toggle('fast');
        return false;
    });
    

    Another way to go:

    $(".show-div").one('click', function() {
        $.post("script.php", { id_1: var_id_1, id_2: var_id_2}, function(data) {
                $("#display").html(data);});
        return false;
    }).click(function() {
        $("#display").toggle('fast');
        return false;
    });​
    
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