问题描述
是否仍然有要向findForPassport添加或传递另外1个变量?
Is There anyway to add or pass 1 more variable to findForPassport ?
在默认的laravel护照登录中,我只能传递2个变量(用户名,密码),但我想再传递1个变量,并在findForPassport中检查该用户是否属于其他表.
In default laravel passport login, i only can pass 2 variable (username, password),but I want to pass 1 more variable and check in findForPassport if that user is belong other table or not .
推荐答案
我希望这个答案可以对其他人有所帮助.
如果要添加变量并将此变量传递给User Authenticate模型中的findPassport函数,则需要更新护照中的3类:
-供应商\联盟\ oauth2-server \ src \ Repositories \ UserRepositoryInterface
中的 UserRepositoryInterface -供应商\联盟\ oauth2-server \ src \ Grant \ PasswordGrant中的 PasswordGrant
-供应商\ laravel \ passport \ src \ Bridge \ UserRepository中的 UserRepository
在示例代码中,我将添加父变量,并且代码将如下所示
I hope this answer can help to other.
If you want to add variable and pass this variable to findPassport function in User Authenticate model , you need to update 3 class in passport :
- UserRepositoryInterface in vendor\league\oauth2-server\src\Repositories\UserRepositoryInterface
- PasswordGrant in vendor\league\oauth2-server\src\Grant\PasswordGrant
- UserRepository in vendor\laravel\passport\src\Bridge\UserRepository
in the example code I will add parent variable and code will look like this
+在UserRepositoryInterface类中
+in UserRepositoryInterface class
interface UserRepositoryInterface extends RepositoryInterface
{
/**
* Get a user entity.
*
* @param string $username
* @param string $password
* @param string $grantType The grant type used
* @param ClientEntityInterface $clientEntity
*
* @return UserEntityInterface
*/
public function getUserEntityByUserCredentials(
$username,
$password,
$parent, <------variable example
$grantType,
ClientEntityInterface $clientEntity
);
}
+在PasswordGrant类中
+in PasswordGrant class
class PasswordGrant extends AbstractGrant{
protected function validateUser(ServerRequestInterface $request, ClientEntityInterface $client)
{
$username = $this->getRequestParameter('username', $request);
if (is_null($username)) {
throw OAuthServerException::invalidRequest('username');
}
$password = $this->getRequestParameter('password', $request);
if (is_null($password)) {
throw OAuthServerException::invalidRequest('password');
}
/**
* Get a user parent.
* varaible example
*/
$parent = $this->getRequestParameter('parent', $request);
if (is_null($parent)) {
throw OAuthServerException::invalidRequest('password');
}
$user = $this->userRepository->getUserEntityByUserCredentials(
$username,
$password,
$parent, <--- variable example get from request
$this->getIdentifier(),
$client
);
if ($user instanceof UserEntityInterface === false) {
$this->getEmitter()->emit(new RequestEvent(RequestEvent::USER_AUTHENTICATION_FAILED, $request));
throw OAuthServerException::invalidCredentials();
}
return $user;
} }
+在UserRepository类中
+in UserRepository class
class UserRepository implements UserRepositoryInterface
{
public function getUserEntityByUserCredentials($username, $password, $parent, $grantType, ClientEntityInterface $clientEntity)
/*add 1more parameter that implement from UserRepositoryInterface*/
{
$provider = config('auth.guards.api.provider');
if (is_null($model = config('auth.providers.'.$provider.'.model'))) {
throw new RuntimeException('Unable to determine authentication model from configuration.');
}
if (method_exists($model, 'findForPassport')) {
$user = (new $model)->findForPassport($username,$parent); <--- finally we pass parent variable to findForPassport here
} else {
$user = (new $model)->where('email', $username)->first();
}
if (! $user) {
return;
} elseif (method_exists($user, 'validateForPassportPasswordGrant')) {
if (! $user->validateForPassportPasswordGrant($password)) {
return;
}
} elseif (! $this->hasher->check($password, $user->getAuthPassword())) {
return;
}
return new User($user->getAuthIdentifier());
}
}
然后您可以从findForPassport的参数中获取$ parent值,但请确保您以有说服力的User返回值.如果要加入表,可以在下面的示例代码中查看
then u can get $parent value from parameter in findForPassport .but make sure you return value as eloquent User .if you want to join table , you can look my example code below
class User extends Authenticatable{
..........
public function findForPassport($identifier,$parent) {
$a = $this
->Join('role as r', 'r.user_id', '=', 'users.id')
->get();
return $a->where('name', $identifier)->where('role_id',$parent)->first();
}
}
这篇关于自定义Laravel护照检查的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!