Php 如何在laravel 5.6中进行自定义登录？

我使用了自己的自定义登录，但没有使用auth。这是我的密码

 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。但它不起作用。当凭证正确时，它会重定向到仪表板，即正确，但当电子邮件、密码或其他凭证错误时，它会显示错误消息。但如果用户输入的数据不在数据库中，则必须显示用户不存在的消息。但它将转到页面未找到错误

---

我想找到这个问题的解决方案。

好的，首先确认您在web中定义了类似“/user”的路由。php

使用以下代码自定义登录。它是用于Ajax登录的，您可以根据自己的需求进行修改

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]);            
    }
}

您没有在

$user

中使用过

get（）

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

它显示

404未找到

错误，因为您使用了

firstorFail（）

它将在没有记录匹配时失败，并将显示

404未找到

错误

$user1 =  User::where('type',$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，因此我建议只使用一个我做了以下更正：

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

检查雄辩查询的计数，如下面的

count（$users->get（））

，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');
}

---

我没有测试代码，但是想法应该很清楚。

您的routeRoute:：match（['get'，'post']，'/user'，'UserController@userLogin');然后检查$user1是否包含任何数据。如果它不返回任何内容或为空，则使用用户不退出的消息重定向。您能否向代码中添加一些解释，以便其他人可以从中学习？