Login with jQuery AJAX is not working

Question

function do_login()
{
$('.loginsubmit').click(function(e)
{
    var datum=$('.loginform').serialize();
    $.ajax({
        type: "POST",
        url: "login",
        data: datum,
        contentType: "application/json; charset=utf-8",
        dataType:"json",
        success: function(data)
        {
            if(data.prop=="false")
            {
                $('p#login-error-notif').text(data.error);
            }
            else if(data.prop=="true")
            {
                $('#logindrp span').text("Welcome "+data.name);
            }
        }
    });            
});
} //closing do_login function

This is the function called when user click on the Login the button. The error I see is the Uncaught reference error: 'prop' of null.

The code in the PHP page is

        if(isset($_POST['username']))
        {
         $username=$_POST['username']

...// proceed with checking the username and password

    if($check_login)
    {
    $return=array(prop=>"true", name=>$name, username=>$username);  
        echo json_encode($return);  
    }
            else
            {
            $return=array(prop=>"false", error=>"Invalid username/password.");  
        echo json_encode($return);
            }


        }

But I observed this part is not being executed. I mean the 'username' is not sent to the page. I checked it with adding the code setcookie("username", $username); inside it. The cookie is never created. What is the problem in my AJAX request?

Obiviously username is not in datum. But what data is? Check with sth like $.each(datum, function(key, value) { alert(key +' => '+ value); }); — AmazingDreams
– AmazingDreams, Commented Aug 8, 2013 at 12:27
have you used var_dump the post and then use the ajax to out put the data on screen to check what values have been submitted? — Liam Sorsby
– Liam Sorsby, Commented Aug 8, 2013 at 12:29
@AmazingDreams: datum is serialized with username and password. Also I passed username and password individually. But no use. — Sarvap Praharanayuthan
– Sarvap Praharanayuthan, Commented Aug 8, 2013 at 12:30

marteljn · Accepted Answer · 2013-08-08 12:47:48Z

1

The contentType property in your AJAX call sets the format of the data sent to the server. By default PHP won't accept JSON. Remove the contentType line from your AJAX call.

By default JQuery sends the data as application/x-www-form-urlencoded; charset=UTF-8, which is likely what your PHP script likely needs to fill the $_POST array properly.

This explains why your username value is not getting set. This is also a very common mistake that people make all of the time.

edited Aug 8, 2013 at 12:47

answered Aug 8, 2013 at 12:39

marteljn

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4 Comments

Sarvap Praharanayuthan Over a year ago

Thank you @marteljn, on removing the line I can see the cookie is getting set. But I am not getting the response back from JSON. Still I am getting this error Uncaught TypeError: Cannot read property 'prop' of null

marteljn Over a year ago

@SuryaS I think you are trying to access it the wrong way. Do a success: function(data){console.log(data) and post the output.

marteljn Over a year ago

@SuryaS it looks like maybe those if blocks are not getting hit meaning that no output is getting outputted as JSON.

Sarvap Praharanayuthan Over a year ago

I have found the error. Seems the prop and error must be enclosed within quotes :) Thank you @marteljn

Sarvap Praharanayuthan · Accepted Answer · 2013-08-08 13:18:37Z

0

It worked well after resolving the following errors:

contentType: "application/json; charset=utf-8", Thank you marteljn.
Enclosing the variables prop, error, username in single codes.

answered Aug 8, 2013 at 13:18

Sarvap Praharanayuthan

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