How to solve Firebug’s “Aborted” messages when sending form data via AJAX?

Question

I have created a PHP web application and use MySQL as database backend, but there is an 'Aborted' note on Firebug's Net panel's status column when I access the web page. Why?

 $('#submit').on('click', function () {
//        e.preventDefault();
        var formData = JSON.stringify($("#frmPayoye").serializeObject());
        console.log(formData);
        $.ajax({
            type: "POST",
 
            url: "http://www.sanaroopay.com/pg/api/ectransact/", 
            data: formData,
            cache: false,
            timeout: 60000,
            async: false,
            processData: true,
            dataType: 'json',   //you may use jsonp for cross origin request
            contentType: "application/json; charset=utf-8",
            crossDomain: true,            
            success: function (data) {
                alert(JSON.parse(data));
//                alert("ok");
                console.log("success");
//                window.location.assign('https://secure.payoye.net/web/merchant');
            },
            error: function () {
                console.log("Failed");
            }
        });
    });

Possible duplicate of [How to solve Firebug’s “Aborted” messages upon Ajax requests?](http://stackoverflow.com/questions/12116641/how-to-solve-firebug-s-aborted-messages-upon-ajax-requests) — Severino Lorilla Jr., Apr 05 '16 at 06:27

score 1 · Answer 1 · answered Apr 05 '16 at 11:30

1

You are not cancelling the form submission so the Ajax call is aborted and the page submits the form as it is designed to do. So you need to stop the form submission.

$('#submit').on('click', function (evt) { 
    evt.preventDefault(); //stop the default action of the button
    //Rest of your code
});

answered Apr 05 '16 at 11:30

epascarello

204,599
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If you have it like the code in your question, you did not define event in the `function(e)` – epascarello Apr 05 '16 at 11:46
$('#submit').on('click', function (evt) { evt.preventDefault(); var formData = JSON.stringify($("#frmPayoye").serializeObject()); console.log(formData); – Rajib Palit Apr 05 '16 at 11:50
Either your updated code is not running do to cache or you have another issue elsewhere. – epascarello Apr 05 '16 at 11:52

score -1 · Answer 2 · edited Jun 20 '20 at 09:12

-1

Please see the documentation of XHR open() for example here: https://developer.mozilla.org/en-US/docs/DOM/XMLHttpRequest

Note: Calling this method an already active request (one for which open()or openRequest()has already been called) is the equivalent of calling abort().

Just create a new XHR instance whenever you need one. Better yet, use jQuery or other JS library to do AJAX. It should shield you from these intricacies.

How to solve Firebug’s “Aborted” messages upon Ajax requests?

edited Jun 20 '20 at 09:12

Community

1
1

answered Apr 05 '16 at 06:28

Severino Lorilla Jr.

1,637
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20
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Tnks for your answer. But i am alredy use ajax. But it still not work ? – Rajib Palit Apr 05 '16 at 09:12
OP is NOT using the same `open()` so that is incorrect. – epascarello Apr 05 '16 at 11:35

How to solve Firebug’s “Aborted” messages when sending form data via AJAX?

2 Answers2