在mysqli Prepared Statement上获取json_encode（）的问题

Question

I am having some problem on exporting the result of a Prepared Statement using MySQLi.

$age = "young";
$rows = array();
$mysqli = new mysqli($db_hostname, $db_username, $db_password, $db_database);

        $stmt = $mysqli->prepare("SELECT fname, lname FROM users WHERE age=?");
        $stmt->bind_param('s', $age);
        $stmt->execute();
        $stmt->bind_result($thefName,$thelName);
                while($stmt->fetch()) 
                  {
                    array_push($rows, $thefName,$thelName);
                  }
            $stmt->close();

$mysqli->close();
echo json_encode($rows);

as you can see I am trying to echo out the json_encode($rows); each of rows as a JSON Array like this :

[
    {"firstName":"John", "lastName":"Doe"}, 
    {"firstName":"Anna", "lastName":"Smith"}, 
    {"firstName":"Peter","lastName":"Jones"}
]

but what I am getting is

["John", "Doe", "Anna", "Smith", "Peter", "Jones"] 

Can you please let me know why this is happening and how I can fix it

Answer 1

Push the proper array -

array_push($rows, array('firstName' => $thefName, 'lastName' => $thelName));