php - 我无法使用 jwt 验证用户

问题描述

我正在尝试使用 Jwt 进行身份验证。找了很多关于Jwt的例子，但是有一个地方我不明白。我可以使用库轻松生成令牌，但我无法验证这里是我使用的库

composer require firebase/php-jwt

我正在创建一个令牌

   // Allow from any origin
if (isset($_SERVER['HTTP_ORIGIN'])) {
    header("Access-Control-Allow-Origin: {$_SERVER['HTTP_ORIGIN']}");
    header('Access-Control-Allow-Credentials: true');
    header('Access-Control-Max-Age: 86400');    // cache for 1 day
}

// Access-Control headers are received during OPTIONS requests
if ($_SERVER['REQUEST_METHOD'] == 'OPTIONS') {

    if (isset($_SERVER['HTTP_ACCESS_CONTROL_REQUEST_METHOD']))
        header("Access-Control-Allow-Methods: GET, POST, OPTIONS");         

    if (isset($_SERVER['HTTP_ACCESS_CONTROL_REQUEST_HEADERS']))
        header("Access-Control-Allow-Headers:        {$_SERVER['HTTP_ACCESS_CONTROL_REQUEST_HEADERS']}");

    exit(0);
}


require_once('vendor/autoload.php');
use \Firebase\JWT\JWT; 
define('SECRET_KEY','Super-Secret-Key'); 
define('ALGORITHM','HS256');  


$postdata = file_get_contents("php://input");
$request = json_decode($postdata);

$action = $_GET['action'];

   if ($action == 'login') {

 $email = 'freaky@jolly.com';
 $password = '12345678';


if($email == "freaky@jolly.com" && $password == "12345678"){

     $iat = time(); // time of token issued at
     $nbf = $iat + 10; //not before in seconds
     $exp = $iat + 60; // expire time of token in seconds

     $token = array(
         "iss" => "http://example.org",
         "aud" => "http://example.com",
         "iat" => $iat,
         "nbf" => $nbf,
         "exp" => $exp,
             "data" => array(
             "id" => 11,
             "email" => $email
         )
     );

    http_response_code(200);

    $jwt = JWT::encode($token, SECRET_KEY);


     $data_insert=array(
         'access_token' => $jwt, 
         'id'   => '007',
         'name' => 'Jolly',
         'time' => time(),
         'username' => 'FreakyJolly', 
         'email' => 'contact@freakyjolly.com', 
         'status' => "success",
         'message' => "Successfully Logged In"
     );


 }else{
     $data_insert=array(
     "data" => "0",
     "status" => "invalid",
     "message" => "Invalid Request"
     ); 
 } 

} else if($action == 'stuff'){

    if (! isset($_SERVER['argv'][1])) {
        exit('Please provide a key to verify');
    }
    $jwt = $_SERVER['argv'][1];

    try {
        $decoded = JWT::decode($jwt, SECRET_KEY, array(ALGORITHM));
        print_r($decoded);

    }catch (Exception $e){

     http_response_code(401);
     $data_insert=array(
     //"data" => $data_from_server,
     "jwt" => $jwt,
     "status" => "error",
     "message" => $e->getMessage()
     );

   } 
}

echo json_encode($data_insert);

创建令牌：http ://www.ismailcakir.com/jwt/test2.php?action=login

{"access_token":"eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJpc3MiOiJodHRwOlwvXC9leGFtcGxlLm9yZyIsImF1ZCI6Imh0dHA6XC9cL2V4YW1wbGUuY29tIiwiaWF0IjoxNTg2NDM4NTg4LCJuYmYiOjE1ODY0Mzg1OTgsImV4cCI6MTU4NjQzODY0OCwiZGF0YSI6eyJpZCI6MTEsImVtYWlsIjoiZnJlYWt5QGpvbGx5LmNvbSJ9fQ.NylRpDkb4cLMxMGAg9ovmLerNf0dLPBHegqmPRDRxlE","id":"007","name":"Jolly","time":1586438588,"username":"FreakyJolly","email":"contact@freakyjolly.com","status":"success","message":"Successfully Logged In"}

东西：http ://www.ismailcakir.com/jwt/test2.php?action=stuff

在这里，我收到 argv 变量未定义的警告，我无法调用您创建的令牌。创建令牌后，我需要能够在不使用 cookie 或会话的情况下调用它。我尝试了很多方法，但我无法理解如何服用。

标签： phpjwt

根据文档：

注意：当 register_argc_argv 被禁用时，此变量不可用。

最有可能是你的情况。您可以在配置中启用它，或者使用$_SERVER我推荐的 superglobal，因为它不受配置差异的影响并且始终有效（包括 CLI）：

$jwt = $_SERVER['argv'][1];

对$argc.

php - 我无法使用 jwt 验证用户

问题描述

解决方案

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