Php 用户登录laravel 5.2时如何获取api_令牌
我正在使用令牌保护来使用post请求中传递的api_令牌对用户进行身份验证。但我无法使用令牌保护检索api_令牌。我想从Php 用户登录laravel 5.2时如何获取api_令牌,php,api,eloquent,laravel-5.2,restful-authentication,Php,Api,Eloquent,Laravel 5.2,Restful Authentication,我正在使用令牌保护来使用post请求中传递的api_令牌对用户进行身份验证。但我无法使用令牌保护检索api_令牌。我想从retrieveByCredentials($credentials)获取用户,并使用validateCredentials($user,$credentials)检查密码,它们位于EloquentUserProvider.php,但要访问它们,我必须使用Auth::guard('api')->provider,其中provider是受保护的属性。所以我必须手动操作。在实现登录
retrieveByCredentials($credentials)
获取用户,并使用validateCredentials($user,$credentials)
检查密码,它们位于EloquentUserProvider.php
,但要访问它们,我必须使用Auth::guard('api')->provider
,其中provider是受保护的属性。所以我必须手动操作。在实现登录时,是否有任何简单的过程来获取api_令牌
AuthController.php中的我的登录方法
public function api_login(Request $request)
{
$validator = Validator::make($request->all(), ['email' => 'required', 'password' => 'required']);
$credentials = $this->getCredentials($request);
$user = Auth::guard('api')->provider->retrieveByCredentials($credentials);
if(is_null($user)) {
return response()->json([
'error' => [
'message' => 'Your email is not registered yet',
'status_code' => 40
]
]);
} elseif(Auth::guard('api')->provider->validateCredentials($user, $credentials)) {
return response()->json([
'success' => [
'api_token' => Auth::guard('api')->user()->api_token,
'message' => 'Login successful',
'status_code' => 200
]
]);
}
return response()->json([
'error' => [
'message' => 'Login failed',
'status_code' => 20
]
]);
}
注意:我使用了provider,但我不能使用它,因为它是一个受保护的属性。在Laravel 5.2中,有两个身份验证保护
'web'
和'api'
,它们分别在SessionGuard
和TokenGuard
中实现,如您所知
TokenGuard
没有用于检查电子邮件和密码验证的模块,它仅使用api\u令牌检查有效性。
但是SessionGuard
<代码>尝试()方法
因此,我们只能将web guard用于登录,并且可以将'auth:api'
中间件用于其他api
这里是登录功能,很容易实现
public function api_login(Request $request)
{
$validator = Validator::make($request->all(), ['email' => 'required', 'password' => 'required']);
$credentials = $this->getCredentials($request);
// only web guard can validate with credentials
$guard = Auth::guard('web');
if ($guard->attempt($credentials)) {
// maybe you can generate api_token again here
return response()->json([
'success' => [
'api_token' => $guard->user()->api_token,
'message' => 'Login successful',
'status_code' => 200
]
]);
}
else {
return response()->json([
'error' => [
'message' => 'Login failed',
'status_code' => 200
]
]);
}
}