Question

我使用了自己的自定义登录名而不使用auth。这是我的代码

 public function userLogin(Request $request){

  if($request->isMethod('post')){
    $data = $request->input();

    $this->validate($request, [
        'uemail'=> 'required|email',
        'user_type'=> 'required',
        'user_id' => 'required',
        'upassword' => 'required|min:6',
      ],
      [
        'uemail.email' => 'Email must be valid',
        'uemail.required' => 'Email is required',
        'user_type.required' => 'Type is required',
        'user_id.required' => 'Name is required',
        'upassword.required' => 'Password is required',
        'upassword.min' => 'Password must be at least 6 characters',
      ]);

      $user_type = $data['user_type'];
      $user_id = $data['user_id'];
      $uemail = $data['uemail'];
      $upassword = $data['upassword'];
      $hashPass = bcrypt($upassword);

      DB::enableQueryLog();

      $user1 =  User::where('type',$user_type)->where('user_id',$user_id)->where('email',$uemail)
      ->where('status',1)->where('deleted_at',null)->firstOrFail();

      $user =  DB::table('users')->where('type',$user_type)->where('user_id',$user_id)->where('email',$uemail)
      ->where('status',1)->where('deleted_at',null);

    //  $query = DB::getQueryLog();
     // $query = end($query);

      $isPasswordCorrect = Hash::check($upassword, $user1->password);

      if($user == null){
        echo "Failed"; die;
      }

       if($user->exists() && $isPasswordCorrect){
          echo "Success"; die;
          Session::put('userSession',$user1->email);
          Session::put('loginSession',$user_type);
          Session::put('idSession',$user1->user_id);

         return redirect('/user/dashboard');
    } else {
       return redirect('/user')->>with('flash_message_error','Invalid Login Credentials..');
    }

  }

    return view('death_notice.user_login');
}

这是我的登录功能。但是它不起作用。凭据正确时，它将重定向到仪表板，即正确，但是，如果电子邮件或密码或其他凭据错误，则会显示错误消息。但是，如果用户输入的数据不在数据库中，则它必须显示一条消息，提示用户不存在。但找不到页面错误...

我想解决这个问题。

Answer 1

好的，首先确认一下，您在web.php中定义了一条类似于“ / user”的路由

Answer 2

您还没有用过get()雄辩的$user

$user =  DB::table('users')->where('type',$user_type)->where('user_id',$user_id)->where('email',$uemail)
  ->where('status',1)->where('deleted_at',null)->get();

它显示404 not found错误，因为您使用过firstorFail()，它将在没有记录匹配时失败，并显示404 not found错误，因此

$user1 =  User::where('type',$user_type)->where('user_id',$user_id)->where('email',$uemail)
  ->where('status',1)->where('deleted_at',null)->first();

if(empty($user1))
{
   echo "user doesn't exists";exit;
}

根据我的理解，查询$user和$user1是相同的，所以我的建议是仅使用一个

Answer 3

我做了以下更正：

$data = $request;就足够了，您不需要$data = $request->input();

检查像这样的count($users->get())口才查询的计数，以检查是否找到了用户。

public function userLogin(Request $request){

    if($request->isMethod('post')){
        $data = $request->input();

        $this->validate($request, [
            'uemail'=> 'required|email',
            'user_type'=> 'required',
            'user_id' => 'required',
            'upassword' => 'required|min:6',
          ],
          [
            'uemail.email' => 'Email must be valid',
            'uemail.required' => 'Email is required',
            'user_type.required' => 'Type is required',
            'user_id.required' => 'Name is required',
            'upassword.required' => 'Password is required',
            'upassword.min' => 'Password must be at least 6 characters',
          ]);

          $user_type = $data['user_type'];
          $user_id = $data['user_id'];
          $uemail = $data['uemail'];
          $upassword = $data['upassword'];
          $hashPass = bcrypt($upassword);

          DB::enableQueryLog();

          $user1 =  User::where('type',$user_type)->where('user_id',$user_id)->where('email',$uemail)
  ->where('status',1)->where('deleted_at',null)->firstOrFail();

          $users =  DB::table('users')->where('type',$user_type)->where('user_id',$user_id)->where('email',$uemail)
  ->where('status',1)->where('deleted_at',null);

//  $query = DB::getQueryLog();
 // $query = end($query);

  $isPasswordCorrect = Hash::check($upassword, $user1->password);

   if(count($users->get()) == 1){
      $user1 = $users->first();
      $isPasswordCorrect = Hash::check($upassword, $user1->password);
      echo "Success"; die;
      Session::put('userSession',$user1->email);
      Session::put('loginSession',$user_type);
      Session::put('idSession',$user1->user_id);

     return redirect('/user/dashboard');
} else {
   return redirect('/user')->>with('flash_message_error','Invalid Login Credentials..');
}

}

return view('death_notice.user_login');
}

我还没有测试代码，但是想法应该很清楚。

Answer 4

使用以下代码自定义登录。用于Ajax登录，您可以根据需要进行修改。

function userLogin(Request $request)
{
    $auth = false;
    $credentials = $request->only('email', 'password');
    // if user login credentials are correct then users logged in 
    if (Auth::attempt($credentials, $request->has('remember')))
    {
        $auth = true; // Success
    }
    // returns json response if login credentials are correct.
    if ($auth == true)
    {
        return response()->json([
            'auth' => $auth,
            'intended' => URL::previous(),
            'msg'=>1,
            'name' =>  Auth::user()->name                
        ]);

    } else { // returns json response if login credentials are incorrect
        return response()->json(['msg' => 2]);            
    }
}

如何在laravel 5.6中进行自定义登录？

4 个答案: