how to console.log result of this ajax call?

Question

I have the following function in jQuery code:

btnLogin.on('click', function(e,errorMessage){
    console.log('my message' + errorMessage);
    e.preventDefault();

    return $.ajax({
        type: 'POST',
        url: 'loginCheck',
        data: $(formLogin).serialize(),
        dataType: 'json'  
    }).promise();
    console.log('my message' + errorMessage);
});

WHAT I AM TRYING TO DO: I am trying to console.log the error message. I am getting undefined if the console.log line is above the ajax function, and nothing if the console.log is bellow of it.

Can anyone tell me how to get the value of the errorMessage displayed in this or another new function?

Also, any link with about using Ajax for checking php login form will be deeply appreciagted

Regards,Zoran

score 15 · Answer 1 · answered Aug 19 '12 at 10:32

15

Why not handle the error within the call?

i.e.

$.ajax({
    type: 'POST',
    url: 'loginCheck',
    data: $(formLogin).serialize(),
    dataType: 'json',
    error: function(req, err){ console.log('my message' + err); }
});

answered Aug 19 '12 at 10:32

Qiau

5,976
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2

it produces undefined...result of the log is my message undefined. – Zoran Aug 19 '12 at 10:34

score 11 · Answer 2 · edited Dec 12 '19 at 15:00

try something like this :

$.ajax({
    type: 'POST',
    url: 'loginCheck',
    data: $(formLogin).serialize(),
    dataType: 'json',
    success: function (textStatus, status) {
        console.log(textStatus);
        console.log(status);
    },
    error: function(xhr, textStatus, error) {
        console.log(xhr.responseText);
        console.log(xhr.statusText);
        console.log(textStatus);
        console.log(error);
    }
});

Nick · Answer 3 · 2012-08-21T10:09:15.477

You could try something like this (copied from the jQuery Ajax examples)

var request = $.ajax({
  url: "script.php",
  type: "POST",
  data: {id : menuId},
  dataType: "html"
});

request.done(function(msg) {
  console.log( msg );
});

request.fail(function(jqXHR, textStatus) {
  console.log( "Request failed: " + textStatus );
});

The problem with your original code is that the error argument you pass into your on function isn't actually coming from anywhere. JQuery on doesn't return a second argument, and even if it did, it would relate to the click event not the Ajax call.

score 7 · Answer 4 · edited Dec 12 '19 at 15:03

7

$.ajax({
    type: 'POST',
    url: 'loginCheck',
    data: $(formLogin).serialize(),    
    success: function(result){
        console.log('my message' + result);
    }
});

edited Dec 12 '19 at 15:03

AndrewRMillar

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answered Jul 10 '13 at 07:18

nsm

337
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6

score 1 · Answer 5 · answered Aug 19 '12 at 12:49

Ajax call error handler will be triggered if the call itself fails.

You are probably trying to get the error from server in case login credentials do not go through. In that case, you need to inspect the server response json object and display appropriate message.

e.preventDefault();
$.ajax(
{
    type: 'POST',
    url: requestURI,
    data: $(formLogin).serialize(),
    dataType: 'json',
    success: function(result){
        if(result.hasError == true)
        {
            if(result.error_code == 'AUTH_FAILURE')
            {
                //wrong password
                console.log('Recieved authentication error');
                $('#login_errors_auth').fadeIn();
            }
            else
            {
                //generic error here
                $('#login_errors_unknown').fadeIn();
            }
        }
    }
});

Here, "result" is the json object returned form the server which could have a structure like:

$return = array(
        'hasError' => !$validPassword,
        'error_code' => 'AUTH_FAILURE'
);
die(jsonEncode($return));

Hi Rohit. Unfortunatly the code produces error in line $.ajax...I keep my first line which uses on click event to fire the code, and place your code in between. Shows error in line with $.ajax — Zoran, Aug 19 '12 at 19:32
Zoran, this code works for me on my demo setup. Can you include the exact error you are seeing or the entire code snippet on the click handler. Please see if the variables 'formLogin' resolve to valid form element and the url option is a valid target url. A much more readable way to get a form data string might be $('#formLoginID').serialize(). — Rohit, Aug 20 '12 at 09:11

score 0 · Answer 6 · answered Jan 10 '18 at 08:34

0

If you want to check your URL. I suppose you are using Chrome. You can go to chrome console and URL will be displayed under "XHR finished loading:"

answered Jan 10 '18 at 08:34

Jeffery

134
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15

score -1 · Answer 7 · answered Mar 16 '14 at 04:20

-1

In Chrome, right click in the console and check 'preserve log on navigation'.

answered Mar 16 '14 at 04:20

weaveoftheride

4,092
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how to console.log result of this ajax call?

7 Answers7

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