PHP/JSON/ANDROID - why my JSONArray getting error - php

Question

i try to displaying value of my table in android, and i put my code into hosting,
but when i try to running my app, my link doesn't work,
and i tried to create new link that displays the same output , and its works fine,
this is my code to encode json
send_data.php
<?php
include 'dbconfig.php';
$con = new mysqli($servername, $username, $password, $dbname);
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = "select id,ask from pertanyaan";
$result = mysqli_query($con, $query);
while ($r = mysqli_fetch_array($result)) {
extract($r);
$rows[] = array(
"id"=>$id,
"ask"=>$ask
);
}
header('Content-Type: application/json');
echo json_encode($rows);
mysqli_close($con);
?>
did I'm doing something wrong??
because I want to set the output from the database

Answer 1

This is just a suggestion without seeing your AJAX code:
Add this code to the top of your PHP code:
header("Access-Control-Allow-Origin: *");
header('Content-type: application/json');
And use this code as an example of AJAX:
var poutput = $('.itemsHolder');
$.ajax({
url: 'https://yourdomain.com/php/YOUR-PAGE.php',
dataType: 'jsonp',
jsonp: 'jsoncallback',
timeout: 5000,
success: function(data, status){
$.each(data, function(pi,item){
str = item.id;
var products = '<div id="'+item.id+'">'+
'</div>';
poutput.append(products);
});
},
error: function(){
//alert('There was an error loading the data.');
}
});
Don't forget to add an element with the class name of itemsHolder to your page.