这是我想在laravel 5.6中进行身份验证的表结构:
权限:Permission_id,权限
登录:login_id,permission_id,全名,密码
电子邮件:email_id,login_id,电子邮件
电话:phone_id,login_id,phone_number
用户可以使用电子邮件和密码登录,也可以使用带密码的电话登录 每个用户可以有许多电话和电子邮件,而我有3种权限类型的用户,admin,manager。
我该怎么做? 这是我的代码,如果电子邮件和电话是登录表中的字段,则可以正常工作 但在我的情况下,电子邮件和电话与登录表分开了,因为每个用户可以拥有多个电子邮件或电话。
<?php
namespace App\Http\Controllers\Auth;
use Illuminate\Http\Request;
use App\Http\Controllers\Controller;
use Illuminate\Foundation\Auth\AuthenticatesUsers;
class LoginController extends Controller {
use AuthenticatesUsers;
public function showLoginForm() {
return view('panel.login');
}
public function username()
{
return 'email';
}
protected function attemptLogin(Request $request)
{
$username = $request->username;
return $this->guard()->attempt(
$this->credentials($request), $request->filled('remember')
);
}
protected $redirectTo = '/panel';
public function __construct() {
$this->middleware('guest')->except('logout');
}
}
?>
答案 0 :(得分:0)
您可以编写此内容。希望这能解决您的问题。
<?php
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use Illuminate\Support\Facades\Input;
use App\User;
use Illuminate\Support\Facades\Auth;
use Illuminate\Foundation\Auth\AuthenticatesUsers;
class LoginController extends Controller {
use AuthenticatesUsers;
/**
* Where to redirect users after login.
*
* @var string
*/
protected $redirectTo = '/panel';
/**
* Create a new controller instance.
*
* @return void
*/
public function __construct() {
$this->middleware('guest', ['except' => 'logout']);
}
/**
* Display a listing of the resource.
*
* @return \Illuminate\Http\Response
*/
public function login() {
return view('panel.login');
}
protected function credentials(Request $request) {
$field = filter_var($request->input($this->username()), FILTER_VALIDATE_EMAIL) ? 'email' : 'username';
$request->merge([$field => $request->input($this->username())]);
return array_merge($request->only($field, 'password'));
}
public function username() {
return 'email';
}
public function authenticate(Request $request) {
$credentials = $this->credentials($request);
if (Auth::attempt($credentials)) {
return redirect()->intended('/');
} else {
return redirect()->intended('login')->with('error', 'Invalid login credentials. Check your email address or username and password!');
}
}
}
答案 1 :(得分:0)
感谢大家的重播
我解决了,这是我的代码
ID是什么,我用login_id和密码进行登录
但是我从电子邮件或电话表中获得了login_id
所以我得到用户输入的电子邮件的login_id
<?php
namespace App\Http\Controllers\Auth;
use Illuminate\Http\Request;
use App\Http\Controllers\Controller;
use Illuminate\Foundation\Auth\AuthenticatesUsers;
use App\Email;
use App\Phone;
class LoginController extends Controller {
use AuthenticatesUsers;
public function showLoginForm() {
return view('panel.login');
}
public function username() {
return 'username';
}
protected function attemptLogin(Request $request) {
$id = '0';
$username=trim( $request->username );
if( filter_var($request->input($this->username()), FILTER_VALIDATE_EMAIL) ) {
$id= Email::select('login_id')->where('email',$username)->first();
}elseif (is_numeric($username)) {
$id= Phone::select('login_id')->where('phone_number',$username)->first();
}
$id=($id==null)?0:$id->login_id;
return $this->guard()->attempt(
['Login_id' => $id, 'password' => $request->password], $request->filled('remember')
);
}
protected $redirectTo = '/panel';
public function __construct() {
$this->middleware('guest')->except('logout');
}
}