Catch status code in jquery.form.js

Question

I'm trying to make a form debugger to my app, and using jquery.form.js to submit the form.

But I want to catch the status code on both success or error, how can I do it?

I tried this, but the sta variable only returns the 'success' or 'error', not the status code I expected, plz help.

$('form').ajaxSubmit({
    success: function(res, sta, xhr, $form) {
        $('#result-code').html(sta);
        $('#result-content').html(res);
    },
    error: function(res, sta, xhr, $form) {
        $('#result-code').html(sta);
        $('#result-content').html(res);
    }
});

as @Rainer Rillke replied: the solution is below, notice that the parameters sequence received of error is differ from the one with success!

$('form').ajaxSubmit({
    success: function(res, sta, xhr, $form) {
        $('#result-code').html(xhr.status);
        $('#result-content').html(res);
    },
    error: function(xhr) {
        $('#result-code').html(xhr.status);
        $('#result-content').html(xhr.responseText);
    }
});

Assuming you are using jquery.form plugIn and want to capture the HTTP server status, use the xhr.status and xhr.statusText properties of the jqXHR. — Rainer Rillke
– Rainer Rillke, Commented Jun 4, 2014 at 1:21
@Rainer Rillke: I tried, but still cannot got it right, I updated the question, plz help. — Alfred Huang
– Alfred Huang, Commented Jun 4, 2014 at 1:30

Rainer Rillke · Accepted Answer · 2014-06-04 01:35:32Z

2

Use the xhr.status property of the jqXHR.

$('form').ajaxSubmit({
    success: function(res, sta, xhr, $form) {
        $('#result-code').text(xhr.status);
        $('#result-content').html(res);
    },
    error: function(xhr) {
        $('#result-code').text(xhr.status);
    }
});

edited Jun 4, 2014 at 1:35

answered Jun 4, 2014 at 1:28

Rainer Rillke

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4 Comments

Alfred Huang Over a year ago

I got undefined on both xhr.status and xhr.statusText.

Rainer Rillke Over a year ago

Updated response. Your parameter order was wrong. Use console.log(arguments) if you are not sure what you get.

Alfred Huang Over a year ago

I got xhr to be a string of value "METHOD NOT ALLOWED" but not an object.

Alfred Huang Over a year ago

Oh, I found the point, the argument list is different, I have update the solution to the question, thank you!

mwilson · Accepted Answer · 2014-06-04 01:09:25Z

0

I'm not sure if it's some kind of custom error that you're trying to catch, but if you're just trying to return the error that is coming from the web page, do a try/catch:

try {
     //do something here
}

catch (e) { //e = exception
     alert(e.toString());
}

answered Jun 4, 2014 at 1:09

mwilson

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2 Comments

Alfred Huang Over a year ago

The procedure don't even raise an error. I found the answer to the standard jquery function here: stackoverflow.com/questions/12759627/…, but I'm using jquery.form.js plugin for ajax submitting the form. I want the solution in this case.

Rainer Rillke Over a year ago

Names of jQuery plugIns aren't unique. Links are cheap.

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