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Home/ Questions/Q 3977904
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T04:58:47+00:00 2026-05-20T04:58:47+00:00

I am trying to make the following recursive function work but keep receiving an

  • 0

I am trying to make the following recursive function work but keep receiving an error (following code):

var checkStatus = function(jobID) {
        $.ajax({
            type: 'post',
            url: '/admin/process/check-encode-status.php',
            data: {jobID: jobID},
            success: function(data) {
                if (data == 'processing') {
                checkStatus(jobID).delay(2000);
                } else {
                    $("#videoOutput").html(data);
                }
            }
        });
     };

The error I am receiving is: checkStatus(jobID) is undefined

Everything seems to be working as it should, Firebug is just throwing this warning. The function is repeating itself, posting the jobID and receiving “processing” from the PHP script.

I had a similar script that I was using that used setTimeout() to be recursive but I could not figure out how to pass the jobID along with calling the function (errors).

Any ideas? Thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-20T04:58:47+00:00Added an answer on May 20, 2026 at 4:58 am

    Just remove the .delay() and you’ll get rid of the error. You should use setTimeout instead.

    var checkStatus = function (jobID) {
        $.ajax({
            type: 'post',
            url: '/admin/process/check-encode-status.php',
            data: {
                jobID: jobID
            },
            success: function (data) {
                if (data == 'processing') {
                    setTimeout(function() { // <-- send an anonymous function
                        checkStatus(jobID); // <--    that calls checkStatus
                    }, 2000);
                } else {
                    $("#videoOutput").html(data);
                }
            }
        });
    };
    

    This is because checkStatus() doesn’t return anything explicitly, so you’re trying to call .delay() on undefined.

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