我有一张mysql表：

mysql> select * from  members;
+-------+-----------+-----------+
| memid | firstname | lastname  |
+-------+-----------+-----------+
|     1 | billal    | begueradj |
|     2 | bill      | gates     |
|     3 | steve     | jobs      |
+-------+-----------+-----------+
3 rows in set (0.00 sec)

我有这个密码：

<?php
$output = array('error' => false);
$members = array();

try {
    $db = new PDO('mysql:host=localhost;dbname=bill;charset=utf8',
                   'root',
                   ''
    );
} catch(Exception $e) {
    die('Error in connecting to DB: <br/>'.$e->getMessage());
}

$response = $db->query('SELECT * FROM members');
while($row = $response->fetch()){
    echo $row['firstname'].' ';
    echo $row['lastname'].'<br/>';
    array_push($members, $row);
}

$output['members'] = $members;
$response->closeCursor();
$json = json_encode($out);
echo $json; // outputs correctly
header("Content-type: application/json");    // error here
die();
?>

当我运行包含上面php代码的php文件时，我收到了这个错误消息：
syntaxerror:json.parse:json数据的第1行第1列出现意外字符
为什么会这样？
p、 当然，当我评论这句话时： //header("Content-type: application/json"); 错误信息消失
编辑：新代码版本如下：

<?php
    header("Content-type: application/json");
    $output = array('error' => false);
    $members = array();

    try {
        $db = new PDO('mysql:host=localhost;dbname=bill;charset=utf8',
                       'root',
                       ''
        );
    } catch(Exception $e) {
        die('Error in connecting to DB: <br/>'.$e->getMessage());
    }

    $response = $db->query('SELECT * FROM members');
    while($row = $response->fetch()){       
        array_push($members, $row);
    }

    $output['members'] = $members;
    $response->closeCursor();
    $json = json_encode($out); 
    //echo $json;    

?>

仍然收到相同的错误消息