无法正确返回JSON数据

Question

Hey I don't use PHP very often maybe once a year and I am having a lot of trouble returning data correctly. Basically I am getting an object that seems to contain my data in the response text which I don't know why. If someone could explain to me why this happens I'd be very grateful!

this is the code from the back end

if(isset($_POST['action']) && !empty($_POST['action'])) {
    switch ($_POST['action']){
        case'getCandidates' :
        {
            $bus = array(
                'latitude' => $row['lat'],
                'longitude' => $row['lng'],
                'icon' => './images/' . $row['busColor'] . '.png'
            );
            array_push($json, $bus);
            $query = "SELECT * from candidates WHERE  status = '$status' AND category = '$category' AND location = '$location'";
            $returnRows = $db->con->query($query);
            if ($returnRows->num_rows > 0) {
                $x = 0;
                // output data of each row
                while($row = $returnRows->fetch_assoc()) {
                    $object = new stdClass();
                    $object->status = $row["status"];
                    $object->first_name = $row["first_name"];
                    $object->last_name = $row["last_name"];
                    $object->category = $row["category"];

                    array_push($aResult, $object);

                }
            } else {
                $aResult[0] = "No results";
            }

//                $aResult['result']  = mysql_fetch_object($returnRows);;
            }

and this is the front end code

returnedCandidates = $.ajax({
    url: "../php/admin.php",
    type: 'POST',
    dataType: 'json',
    data: {action: 'getCandidates'},
    success: function(data, textStatus, jqXHR) {

        alert(data);
    }});
console.log(JSON.parse(returnedCandidates[0]));

and this is the line for returning data with php. I forgot to add it.

print_r(json_encode ($aResult));

Answer 1

comment line:

//print_r(json_encode ($aResult));

and return like this:

return (json_encode ($aResult));

Answer 2

Don't use print_r for return.

Try this code:

header('Content-Type: application/json');
echo json_encode($aResult);