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Php 尝试获取非对象laravel 5的属性-我的错误是什么?_Php_Laravel_Laravel 5 - Fatal编程技术网

Php 尝试获取非对象laravel 5的属性-我的错误是什么?

Php 尝试获取非对象laravel 5的属性-我的错误是什么?,php,laravel,laravel-5,Php,Laravel,Laravel 5,我犯了什么错 雇员控制员: public function index() { $employees = Employee::all(); return view('employee.view_all_employee_details', compact('employees')); } 员工模型功能: protected $primaryKey = "emp_id"; protected $with = 'additionaldetails'; public function add

我犯了什么错

雇员控制员:

public function index()
{
  $employees = Employee::all();
  return view('employee.view_all_employee_details', compact('employees'));
}
员工模型功能:

protected $primaryKey = "emp_id";
protected $with = 'additionaldetails';
public function additionaldetails()
{
 return $this->hasOne('App/EmployeeAdditionalDetail' , 'emp_id' , 'emp_id');
}
员工其他详细信息模型:

public function employeeDetails()
{
 return $this->belongsTo('App\Employee' , 'emp_id' , 'emp_id');
}
查看所有员工详细信息视图

@if(count($employees) > 0)

@foreach($employees as $employee)
    <tr>
        <td> {{ $employee->first_name }} </td>
        <td> {{ $employee->manager_id }} </td>
        <td>

          {{ $employee->additionaldetails->emp_id }}  **Error showing here**
        </td>
    </tr>
@endforeach

@else
   {!! "<tr><td>No Recod Found</td></tr>" !!}

@endif
@if(计数($employees)>0)
@foreach($employees作为$employees)
{{$employee->first_name}
{{$employee->manager\u id}
{{$employee->additionaldetails->emp_id}}**此处显示错误**
@endforeach
@否则
{!!“未找到记录”!!}
@恩迪夫
我的错误:

正在尝试获取非对象的属性

试试这个

{{ $employee->additionaldetails->emp_id }}
员工模型可以访问basicdetails()函数。您正在尝试使用additionaldetails()获取数据

也改变这一行

return $this->hasOne('App/EmployeeAdditionalDetail' , 'emp_id' , 'emp_id');


同样在员工模型
protected$with='additionaldetails'中应该受
保护$with=array('basicdetails')谢谢shuvrow。我已更改为{{$employee->basicdetails->emp_id}。但还是一样的问题。@Slow loris。谢谢。但仍然面临同样的问题(plz提供$employee->additionaldetails@sankaran请检查此答案和您的问题的编辑,相应地修改您的代码,然后重试,它应该可以工作。
return $this->hasOne('App\EmployeeAdditionalDetail' , 'emp_id' , 'emp_id');