我必须用我的雄辩来用户模型:
- 用户
- 办公室用户
OfficeUser 在 JWT 配置中定义为标准型号。现在我已经编写了一个中间件来验证它们中的每一个
身份验证用户:
public function handle($request, Closure $next)
{
Config::set('auth.providers.users.model', 'App'User::class);
try {
if (! $user = JWTAuth::parseToken()->authenticate()) {
return response()->json(['user_not_found'], 404);
}
} catch (Tymon'JWTAuth'Exceptions'TokenExpiredException $e) {
return response()->json(['token_expired'], $e->getStatusCode());
} catch (Tymon'JWTAuth'Exceptions'TokenInvalidException $e) {
return response()->json(['token_invalid'], $e->getStatusCode());
} catch (Tymon'JWTAuth'Exceptions'JWTException $e) {
return response()->json(['token_absent'], $e->getStatusCode());
}
return $next($request);
}
身份验证办公室用户
public function handle($request, Closure $next)
{
try {
if (! $user = JWTAuth::parseToken()->authenticate()) {
return response()->json(['user_not_found'], 404);
}
} catch (Tymon'JWTAuth'Exceptions'TokenExpiredException $e) {
return response()->json(['token_expired'], $e->getStatusCode());
} catch (Tymon'JWTAuth'Exceptions'TokenInvalidException $e) {
return response()->json(['token_invalid'], $e->getStatusCode());
} catch (Tymon'JWTAuth'Exceptions'JWTException $e) {
return response()->json(['token_absent'], $e->getStatusCode());
}
return $next($request);
}
此外,我每个人都有一个登录功能:
登录用户
if ($user){
if (Hash::check($request->password, $user->password)) {
// grab credentials from the request
$credentials = $request->only('email', 'password');
try {
// attempt to verify the credentials and create a token for the user
Config::set('auth.providers.users.model', 'App'User::class);
if (! $token = JWTAuth::attempt($credentials)) {
return response()->json(['error' => 'invalid_credentials'], 401);
}
} catch (JWTException $e) {
// something went wrong whilst attempting to encode the token
return response()->json(['error' => 'could_not_create_token'], 500);
}
登录办公室用户
if ($user){
if (Hash::check($request->password, $user->password)) {
// grab credentials from the request
$credentials = $request->only('email', 'password');
try {
// attempt to verify the credentials and create a token for the user
Config::set('auth.providers.users.model', 'App'OfficeUser::class);
if (! $token = JWTAuth::attempt($credentials)) {
return response()->json(['error' => 'invalid_credentials'], 401);
}
} catch (JWTException $e) {
// something went wrong whilst attempting to encode the token
return response()->json(['error' => 'could_not_create_token'], 500);
}
不幸的是,当我登录并尝试调用authUser中间件后面的路由时,我得到了一个"user_not_found"
有人知道为什么会这样吗?办公室用户身份验证工作正常
为发现此问题的任何人发帖
虽然不建议有两个用户表,但我有一个类似的要求,即与我们的一个客户端设置 JWT。这就是我解决问题的方式。
无需在"配置/身份验证.php中对提供程序进行任何更改
'providers' => [
'users' => [
'driver' => 'eloquent',
'model' => App'User::class,
],
]
在身份验证控制器中,通过设置
'Config::set('auth.providers.users.model', 'App'Trainer::class);
示例代码
In authenticate() 方法
if ($credentials['user_type'] == 'consultant') {
'Config::set('auth.providers.users.model', 'App'Trainer::class);
} else {
'Config::set('auth.providers.users.model', 'App'User::class);
}
//Find the user
//Create the token
if ($user) {
$customClaims = ['user_type' => $credentials['user_type']];
$token = JWTAuth::fromUser($user,$customClaims);
} else {
return response()->json(['error' => 'invalid_credentials'], 401);
}
在解析令牌以对用户进行身份验证时,还必须执行相同的操作。示例代码
在 getAuthenticatedUser() 方法中
$payload = JWTAuth::parseToken()->getPayload();
$user_type = $payload->get('user_type');
if($user_type === 'consultant'){
'Config::set('auth.providers.users.model', 'App'Trainer::class);
}else{
'Config::set('auth.providers.users.model', 'App'User::class);
}
if (!$user = JWTAuth::parseToken()->authenticate()) {
return response()->json(['user_not_found'], 404);
}
您可以
更改每个控制器中的__construct
函数,如下所示。以便 jwt 知道要对哪个模型进行身份验证。
办公室用户控制器
function __construct()
{
Config::set('jwt.user', OfficeUser::class);
Config::set('auth.providers', ['users' => [
'driver' => 'eloquent',
'model' => OfficeUser::class,
]]);
}