jQuery $.getJSON not returning data

Question

I am making an AJAX call to my PHP script, via $.getJSON. The problem is, PHP is returning json-encoded data, but my script is not console.logging anything at all.

function Club(){
    this.URL = "http://localhost/imp03/includes/ajaxCall.php";
}
Club.prototype.loadData = function(){
    $.getJSON(Club.URL, function(data){
        console.log(data);
    });
}

There are no errors in the console, and this is my PHP script.

$db = new Database(HOST, USER, PASS, DB);
$array = $db->getData();
header('Content-Type: application/json');
echo json_encode($array);
exit;

This is the method in my Database class, responsible for getting data

public function getData(){
    try{
        $sql = $this->db->prepare("SELECT * FROM inlever3");
        $sql->execute();
        $result = $sql->fetchAll();
        return $result;
    } catch(PDOException $e){
        echo $e;
        return false;
    }
}

first of all, $.getJSON need server to return json type data, json_encode(array) is needed — TechStone
– TechStone, Commented Apr 25, 2014 at 9:49
What do you see when you navigate to http://localhost/imp03/includes/ajaxCall.php in your browser ? — Holt
– Holt, Commented Apr 25, 2014 at 9:50
@TechStone I have edited my code, included the header and I am json_encoding my data, still no change. — user3239713
– user3239713, Commented Apr 25, 2014 at 9:52
@Dave: It is hard to include a jsFiddle, as I can not include PHP. Holt: I am seeing the JSON data I am asking for. ;) — user3239713
– user3239713, Commented Apr 25, 2014 at 9:53

xdazz · Accepted Answer · 2014-04-25 09:52:22Z

3

Club is the function, which doesn't have the URL property.

Club.prototype.loadData = function(){
    $.getJSON(Club.URL, function(data){
        console.log(data);
    });
}

should be

Club.prototype.loadData = function(){
    // here change Club.URL to this.URL
    $.getJSON(this.URL, function(data){
        console.log(data);
    });
}

And use it like:

var club = new Club();
club.loadData();

answered Apr 25, 2014 at 9:52

xdazz

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Comments

Latheesan · Accepted Answer · 2014-04-25 09:53:21Z

2

Just tried it, it works as intended:

You need to use this.URL not Club.URL on the $.getJSON() call

test.php

<?php

if (isset($_GET['loadData']))
{
    // dummy data
    exit(json_encode(
        array('hello' => 'world')
    ));
}

?>

<script type="text/javascript" src="http://code.jquery.com/jquery-1.11.0.min.js"></script>
<script type="text/javascript">

function Club(){
    this.URL = "test.php?loadData";
}
Club.prototype.loadData = function(){
    $.getJSON(this.URL, function(data){
        console.log(data);
    });
}

var c = new Club();
c.loadData();

</script>

answered Apr 25, 2014 at 9:53

Latheesan

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4 Comments

vimal1083 Over a year ago

It wont make any change !!

Latheesan Over a year ago

@Vimalraj.S what do you mean?

vimal1083 Over a year ago

there might be problem in his URL

vimal1083 Over a year ago

@LatheesanKanes OOps sorry , here he could even use c.URL

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