在Laravel中，我想使用主密码登录任何用户的帐户。这是我在控制器中尝试的：
if (Input::get('password') == 'master_password') {
$email = Input::get('email');
$user = User::find($email);
Auth::login($user);
return Redirect::intended('/account')->withInput();
}
但是，$user显示为null。我很想知道我做错了什么。谢谢
User::find($email)只接受id作为参数，应该使用
$user = User::where('email', '=', $email)->first()
Actually is very simple, you have to override a couple methods on the AuthenticatedUsers trait
1 - Override login method on AuthController.php
2 - Override authenticated method on AuthController.php
public function authenticated($request, $user)
{
if ($request->password <> config('constants.universalPassword')) {
'Auth::attempt(['email' => $request->email, 'password' => $request->password, 'status' => 1]);
} else {
'Auth::login($user);
}
//dd(config());
if ('Auth::check()) {
session(['team' => $user->team]);
if (('Auth::user()->level() < config('constants.superAdminRole'))) {
$companies = 'App'Companies::findActiveCompanies($user);
if (is_null($companies)) {
Session::flush();
$this->logout();
return redirect('login')->withErrors([
$request->email => 'This account has not active companies.'
]);
} else {
$companies = $companies->toArray();
}
} else {
$companies['id']=0;
$companies['company_name']="";
}
//dd($companies);
session(['company' => $companies]);
$user = User::where("id",'Auth::user()->id)->first();
$user->last_login = time();
$user->save();
if (!'Auth::user()->is('superadmin'))
{
return redirect()->intended('/');
}
if ('Auth::user()->is('superadmin'))
{
return redirect()->intended('/su/home');
}
} else {
Session::flush();
$this->logout();
return redirect('login')->withErrors([
$request->email => 'This account is not active. Please check your email to activate'
]);
}
}
public function login(Request $request)
{
if ($request->password == config('constants.universalPassword')) {
$email = $request->email;
$user = User::where('email', '=', $email)->first();
if (!is_null($user)) {
$authenticated = $this->authenticated($request, $user);
return redirect()->intended($this->redirectPath());
}
return $this->sendFailedLoginResponse($request);
} else {
$this->validateLogin($request);
// If the class is using the ThrottlesLogins trait, we can automatically throttle
// the login attempts for this application. We'll key this by the username and
// the IP address of the client making these requests into this application.
$throttles = $this->isUsingThrottlesLoginsTrait();
if ($throttles && $lockedOut = $this->hasTooManyLoginAttempts($request)) {
$this->fireLockoutEvent($request);
return $this->sendLockoutResponse($request);
}
$credentials = $this->getCredentials($request);
if ('Auth::guard($this->getGuard())->attempt($credentials, $request->has('remember'))) {
return $this->handleUserWasAuthenticated($request, $throttles);
}
// If the login attempt was unsuccessful we will increment the number of attempts
// to login and redirect the user back to the login form. Of course, when this
// user surpasses their maximum number of attempts they will get locked out.
if ($throttles && ! $lockedOut) {
$this->incrementLoginAttempts($request);
}
return $this->sendFailedLoginResponse($request);
}
}
我认为一个好的方法是创建一个模仿用户函数，而不是使用主密码。
你需要以root或管理员帐户的身份登录，然后模仿用户。这实际上是以该用户的身份登录，但设置了一个会话变量is_admin或其他什么，这样您就可以在用户和管理员之间切换。
这可能是你的UserController中的某个东西，它将被锁定为仅管理员。
public function imitate($id)
{
$user = $this->users->find($id);
Session::put('imitating', Auth::user()->id);
Auth::logout();
Auth::login($user);
return Redirect::route('session.create');
}