CODEWITHSUNDEEP
javascriptphpjqueryajaxjson
Joe Saad
Joe Saad

Reputation: 1950

jQuery[number] was not called

Ok, I have searched this question and have found numerous answers on it but none have worked for me. I'm getting jQuery21006460414978209883_1395689439888 was not called error. The AJAX call is done right, status is 200 and the php url that i'm trying to call, i have validated its JSON and it is a valid json according to jsonlint. I'm using jsonp since it is a crossdomain call. Any advise would help me debug this. Here is my javascript code: 

$.ajax({
    crossDomain: true,
    cache: false,
    type: "GET",
    url: "http://example.com/backend.php",
    data: "fdaf",
    //async: false,
    dataType: "jsonp",
    contentType: 'application/json; charset=utf-8',
    success: function(msg){
      alert("success"+ JSON.stringify(msg));
    },
    error:function (xhr, ajaxOptions, thrownError){
                alert('error function status is : '+xhr.status);
                alert('error thrown is:  '+thrownError);
    },
    jsonp: "callback",
    complete: function (requestState) {
        alert("working");
    } 
});

My backend.php code is here: 

<?php 
header('Content-Type: application/json');

$arr = array('a' => "1", 'b' => "2", 'c' => "3", 'd' => "4");
$tr =  json_encode($arr);

echo $_GET['callback'].'['.$tr.']';
?>

Please advise what I am doing wrong here. Many thanks!

Upvotes: 1

Views: 914

Answers (1)

Niels
Niels

Reputation: 49919

Your issue is this, the output of your PHP:

callbackfunction[[DATA]]

This does not call a Javascript function. You should have this output:

callbackfunction([DATA])

With this PHP:

echo $_GET['callback'].'('.$tr.')';

Because $tr is an Array it will return [DATA]

You should only need JSONP if you do a Cross-Domain request. Otherwise just use a GET / POST.

Upvotes: 2

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