PHP Slim解码JWT令牌

Question

我是PHP新手，使用jwt授权令牌开发restful服务。我已经关注this GitHub example 并且已经在某种程度上理解了代码，但我在这条线路上遇到错误$stmt->bindParam("user_id", $decoded->context->user->user_id);说

注意：在C：\ xampp \ htdocs \ slim2 \ src \ routes.php中的数组转换为字符串。

请帮我解决这个问题，我无法理解什么是context-> user-> user_id来自哪里。完整代码如下

// The route to get a secured data.
$app->get('/restricted', function (Request $request, Response $response) {

$jwt = $request->getHeaders();

$key = "testsecretekey";

try {
    $decoded = JWT::decode($jwt['HTTP_AUTHORIZATION'][0], $key, array('HS256'));
} catch (UnexpectedValueException $e) {
    echo $e->getMessage();
}

if (isset($decoded)) {
    $sql = "SELECT * FROM tokens WHERE user_id = :user_id";

    try {
        $db = $this->db;
        $stmt = $db->prepare($sql);
        $stmt->bindParam("user_id", $decoded->context->user->user_id);
        $stmt->execute();
        $user_from_db = $stmt->fetchObject();
        $db = null;

        if (isset($user_from_db->user_id)) {
            echo json_encode([
                "response" => "This is your secure resource !"
            ]);
        }
    } catch (PDOException $e) {
        echo '{"error":{"text":' . $e->getMessage() . '}}';
    }
  }
});

Answer 1

您只需将令牌发送到JWT：decode。将您的代码更改为：

$jwt = str_replace('Bearer ', '', $jwt['HTTP_AUTHORIZATION'][0]);
$decoded = JWT::decode($jwt, $key, ['HS256']);