问题描述
请帮助我。我是Laravel的新手。 我正在尝试从表中查看数据,但显示此错误。我已经在控制器中定义了模型。我将index.blade.PHP放入文件中。 这些是我的代码。我使用Laravel 7
应用-> User.PHP
<?PHP
namespace App;
use Illuminate\Database\Eloquent\Model;
class User extends Model
{
protected $table = "users";
}
应用-> Http->控制器-> AdminController.PHP
<?PHP
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use App\User;
class AdminController extends Controller
{
public function user(Request $request){
return view('Admin/Form User/index');
}
public function show()
{
$users = User::all();
return view('index',['index' => $users]);
}
}
路线-> web.PHP
Route::get('/Admin/Form User/index','AdminController@user');
Route::get('/index','AdminController@show');
资源->视图-> main.blade.PHP
<a href="Admin/Form User/index" class="btn btn-primary" role="button">Form User</a>
资源->视图->管理员->表单用户-> index.blade.PHP
<table class="table table-bordered table-hover table-striped">
<thead>
<tr>
<th>Id</th>
<th>Class</th>
<th>Name</th>
<th>Option</th>
</tr>
</thead>
<tbody>
@foreach($users as $u)
<tr>
<td>{{ $u->class }}</td>
<td>{{ $u->name }}</td>
<td>
<a>Edit</a>
<a>Delete</a>
</td>
</tr>
@endforeach
</tbody>
</table>
谢谢
解决方法
您应该尝试在用户模型文件中使用这种方式。我认为这是可行的
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
class User extends Model
{
protected $fillable = [
'name','class',];
}
,
您应该尝试以这种方式在AdminController中传递对象,
public function show()
{
$users = User::all();
return view('index')->with('users',$users);
}