我必须用我的 Eloquent 来表达用户模型:
- 用户
- 办公用户
OfficeUser 在 JWT 配置中定义为标准模型。 现在我已经编写了一个中间件来验证它们中的每一个
授权用户:
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);
}
authOfficeUser
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,我为他们每个人提供了一个登录功能:
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 $e) {
// something went wrong whilst attempting to encode the token
return response()->json(['error' => 'could_not_create_token'], 500);
}
登录Office用户
if ($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 $e) {
// something went wrong whilst attempting to encode the token
return response()->json(['error' => 'could_not_create_token'], 500);
}
不幸的是,当我登录并尝试调用 authUser 中间件后面的路由时,我得到一个“user_not_found”
有人知道为什么会这样吗? OfficeUser 身份验证工作正常
最佳答案
发帖给任何发现此问题的人
虽然不建议有两个用户表,但我有一个类似的要求,即与我们的一个客户一起设置 JWT。这就是我解决问题的方法。
无需对 `config/auth.php' 中的提供程序进行任何更改
'providers' => [
'users' => [
'driver' => 'eloquent',
'model' => App\User::class,
],
]
在您的身份验证 Controller 中,通过设置动态修改提供者使用的模型
\Config::set('auth.providers.users.model',\App\Trainer::class);
示例代码
在 authenticate() 方法中
if ($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 token 以对用户进行身份验证时,您也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);
}
关于php - 多用户模型 Laravel JWT Auth,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36839158/