Facade\Ignition\Exceptions\ViewException试图获取属性'；名称'；index.blade.php中非对象的

我有两个表：部门和员工。我想列出部门或员工。部门可以，但当我想看到员工时，它会给我这个错误。我尝试了一些解决方案，但它没有帮助我，它仍然显示这个错误。

这是employees\index.blade.php

----
                <table class="table">
                    <tr>
                        <td>Name</td>
                        <td>Department</td>
                    </tr>
                    @foreach($employees as $employee)
                    <tr>
                        <td>{{ $employee->name }}</td>
                        <td>{{  $employee->department->name }}</td>
                    </tr>
                    @endforeach
                </table>
---

departments/index.blade.php

...
                <table class="table">
                    <tr>
                        <td>Name</td>
                    </tr>
                    @foreach($departments as $department)
                    <tr>
                        <td>{{ $department->name }}</td>
                    </tr>
                    @endforeach
                </table>
...

部门控制员

   public function index()
    {
        $employees = \App\Employee::orderBy('id', 'desc')->paginate(10);

        return view('employees.index')->with('employees', $employees);
    }

    public function index()
    {
        $departments = \App\Department::all();
        $departments = \App\Department::orderBy('id', 'desc')->get();
        $departments = \App\Department::orderBy('id', 'desc')->paginate(5);
        return view('departments.index')->with('departments', $departments);
    }

我得到的错误是 外观\点火\异常\视图异常 试图获取非对象的属性“name”（视图：C:\xampp\htdocs\Laravel\test2\resources\views\employees\index.blade.php）

如果我返回\App\Employee:：orderBy（'id'，'desc'）->paginate（10）；我明白了

{"current_page":1,"data":[{"id":2,"name":"Laura","department":""},{"id":1,"name":"Ana","department":""}],"first_page_url":"http:\/\/localhost\/Laravel\/test2\/public\/employees?page=1","from":1,"last_page":1,"last_page_url":"http:\/\/localhost\/Laravel\/test2\/public\/employees?page=1","next_page_url":null,"path":"http:\/\/localhost\/Laravel\/test2\/public\/employees","per_page":10,"prev_page_url":null,"to":2,"total":2}

---

因此，我将假设这种关系是一个部门有许多员工。这就是设置模型关系的方式

---

所以我假设这个关系是一个部门有很多员工。这就是设置模型关系的方式

u可以使用{{$employee->department->name？？？}来消除此错误。
u可以使用{{$employee->department->name？？？}来消除此错误。
illumb\Database\elounce\Collection{307▼   #项目：阵列：2[▼     0=>App\Employee{308▶}     1=>App\Employee{#309▶}   ] }照亮\数据库\雄辩\收藏{#307▼   #项目：阵列：2[▼     0=>App\Employee{308▶}     1=>App\Employee{#309▶}   ] }酷，这看起来不错，你能做
\App\Employee:：with（'department'）->orderBy（'id'，'desc'）->paginate（10）；
它给我这个错误：lluminate\Database\elounce\RelationNotFoundException调用模型[App\Employee]上未定义的关系[department]。好的，这意味着
部门
和
员工
之间没有关系，您在这些表之间设置了外键吗？非常感谢，我正试图查看它，这就是问题所在。问题解决了。我没有添加关系，这就是原因。酷，这看起来不错，您能用（'department'）执行
\App\employee:：吗->orderBy（'id'，'desc'）->paginate（10）；
它给我这个错误：lluminate\Database\elounce\relationnotfound调用模型[App\Employee]上未定义的关系[department]。好的，这意味着
部门
和
员工
之间没有关系，您是否在这些表之间设置了外键？非常感谢，我正在尝试查看它，这就是问题所在。问题已经解决。我没有添加关系这就是原因。
{"current_page":1,"data":[{"id":2,"name":"Laura","department":""},{"id":1,"name":"Ana","department":""}],"first_page_url":"http:\/\/localhost\/Laravel\/test2\/public\/employees?page=1","from":1,"last_page":1,"last_page_url":"http:\/\/localhost\/Laravel\/test2\/public\/employees?page=1","next_page_url":null,"path":"http:\/\/localhost\/Laravel\/test2\/public\/employees","per_page":10,"prev_page_url":null,"to":2,"total":2}