jQuery ajax JSON custom error handling

Question

I want to working with login system jQuery ajax and json. I'm facing no problem with success function. I dn't know how to through custom errors like "incorrect login".

Here is my code, Please help me.

<script>
    $(document).ready(function(){
        $("#login").click(function(){
            if($("#defaultEmail").val() == ""){
                $("#defaultEmail").addClass("field-err");
                $("#defaultEmail").focus();
            }else if($("#defaultPassword").val() == ""){
                $("#defaultEmail").removeClass("field-err");
                $("#defaultPassword").addClass("field-err");
                $("#defaultPassword").focus();
            }else{
                $("#defaultPassword").removeClass("field-err");
                $('#loading').css('display', 'block');
                var email = $('#defaultEmail').val();
                var password = $('#defaultPassword').val();
                var queryString = "email=" + email + "&password=" + password + "";
                $.ajax({
                  url: 'secureLogin.php',
                  contentType: "application/json; charset=utf-8",
                  dataType:'json',
                  async: false,
                  type:'GET',
                  data: queryString,
                  success: function(data){
                    $('#loading').css('display', 'none');
                    var url = "c_profile.php";
                    $(location).attr('href',url);
                  },
                  error: function(request,error){
                    $('#loading').css('display', 'block');
                    $("#log-ajax-error").append("ERROR: " + error);
                  }
                });
            }
        });
    });
</script>

Here is my sucureLogin.php code

$email = $_GET['email'];
$password = $_GET['password'];
$encrypted_pass = md5($password);

    $sel = @mysql_query("SELECT `name`, `company`, `cid` FROM `company` WHERE `email` = '$email' AND `password` = '$encrypted_pass'");
    $count = @mysql_num_rows($sel);
    if($count > 0){
        while($data = @mysql_fetch_array($sel)){
                       // Success here
        }
    }else{
        $arr = array('a' => 'Unable to login');
        echo json_encode($arr);
    }

i'm throwing json_encode for the error. Please help me with this. Thanks in advance.

aziz punjani · Accepted Answer · 2012-10-01 13:36:52Z

3

On the server side

$arr = array('error' => 'Unable to login');
echo json_encode($arr);

On the client side

success: function(data){
    $('#loading').hide(); 
    if( 'error' in data ){ 
        $("#log-ajax-error").append("ERROR: " + data.error);
    }
    else{
       var url = "c_profile.php";
       $(location).attr('href',url);
    }
},

The error callback will be called if there's an error with the ajax request, so your code needs to be in the success handler.

answered Oct 1, 2012 at 13:36

aziz punjani

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2 Comments

Radhakrishna Chowdary Over a year ago

Dear aziz.punjani, Every thing is fine. Its throwing an error if i give false data. Problem is After throwing false data if you try again with correct login its not redirecting. i.e its always going to the if loop. Please help

Radhakrishna Chowdary Over a year ago

$arr = array('success' => 'Please wait while we redirecting..'); echo json_encode($arr); Ohh i got it :) Thank you so much

Alex · Accepted Answer · 2012-10-01 13:36:26Z

1

In ajax, the error handler does not constitute an error thrown by you, it constitutes a connection error, or an error with the request. This means that, no matter how you send back an error response like you have, it will hit the success callback.

In your success you should do some checks for the response:

success: function(data){

   if(data.status === false)
   {

      // There was an error

   } else {

      $('#loading').css('display', 'none');
      var url = "c_profile.php";
      $(location).attr('href',url);

   }

},

And in your PHP handler add a status key:

$arr = array(
   'status' => true (or false),
   'a' => 'Unable to login'
);

answered Oct 1, 2012 at 13:36

Alex

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