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Home/ Questions/Q 8523667
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T07:28:30+00:00 2026-06-11T07:28:30+00:00

I have a native JavaScript object, that I would like to assign the .ajaxSuccess

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I have a native JavaScript object, that I would like to assign the .ajaxSuccess callback to. The purpose of this is because I want my data model to update after an ajax call succeeds, but I don’t want to make my data model global to the entire JavaScript file. And yes, I checked to make sure my jQuery is included before my script file.

Here is the code:

$("#formButtonAddLink").click(function() {
    $.ajax({
        type: "POST",
        url: "ajax/addlink",
        data: {content: $("#formInputLinkContent").val(), subject: $("#formInputLinkSubject").val()},
        dataType: "json",
        error: function() {
            alert("An ajax error occured adding link")
        }
    });
    return false; //prevents html form submit
})

$(document).ready(function(){
    var links = new Links(20,0);
    $(links).ajaxSuccess(function() {
        console.log("Hey.")    //This does not work.
    });
    $(document).ajaxSuccess(function() {
        console.log("Document hey.")   //This shows up.
    });
});
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T07:28:31+00:00Added an answer on June 11, 2026 at 7:28 am

    The basic idea of ajaxSuccess() is:

    Whenever an Ajax request completes successfully, jQuery triggers the
    ajaxSuccess event. Any and all handlers that have been registered with
    the .ajaxSuccess() method are executed at this time.

    As far as I understand it from the documentation you can only attach the ajaxSuccess() event handler to an element, not a JavaScript object. That is why it works when attaching the event handler to the document.

    We can attach our event handler to any element.

    In this case if you have something similar to:

    <div class="links"></div>
    

    You could do:

    $(document).ready(function(){
        $(".links").ajaxSuccess(function() {
            console.log("Hey.")
            // use links in here:
            var links = new Links(20,0);
        });
    });
    
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