Php 尝试获取非对象laravel 5的属性-我的错误是什么?
我犯了什么错 雇员控制员: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
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 $this->hasOne('App/EmployeeAdditionalDetail' , 'emp_id' , 'emp_id');
到
同样在员工模型
protected$with='additionaldetails'中代码>应该受保护$with=array('basicdetails')代码>谢谢shuvrow。我已更改为{{$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');