在 laravel 中执行登录操作时如何解决下面提到的错误

Question

在下面找到控制器代码：

class SigninController extends Controller
{
  protected function signin(){
    $login = Whmcs::validatelogin(array(
        'email' => Input::get('email'),
         'password2' => Input::get('password2'),
    ));

    if($login->result == 'success') {
        echo 'User Logged In';
    } elseif($login->result == 'error') {
        echo $login->message;
    }
  }
}

路由代码如下：

Route::post('signin','SigninController@signin')->name('login.signin');

blade 文件中的表单操作如下：

 <form class="bs-example form-horizontal" action="{{route('login.signin')}}" method="post">
      <input type = "hidden" name = "_token" value = "<?php echo csrf_token(); ?>">

        <div class="form-group">
          <label class="col-lg-2 control-label">Email</label>
          <div class="col-lg-6">
            <input type="email" class="form-control" name="email" placeholder="Email">
          </div>
        </div>

        <div class="form-group">
          <label class="col-lg-2 control-label">password</label>
          <div class="col-lg-6">
            <input type="password" class="form-control" name="password2" placeholder="password">
          </div>
        </div>
        <div class="form-group">
          <div class="col-lg-offset-2 col-lg-10">
            <button type="submit" class="btn btn-sm btn-success">Submit</button>
          </div>
        </div>

      </form>

在执行上面的代码时，我收到一个错误

Trying to get property 'result' of non-object. Suggest me solution to solve this while performing login action.

Answer 1

尝试使用

回显变量 $login

dd($login);

很可能 validatelogin 函数的输出不是一个对象。