Question

我编写了一个从数据库中删除记录的Controller，这非常简单。代码如下所示。

public function destroy(Request $request)
    {
        try {
            MyModel::where('id', json_decode($request->get('data'),true)['id'])->delete();
            return response()->json([
                'status' => 'success',
                'message' => 'Deleted successfully'
            ]);
        } catch (\Exception $e) {
            return response()->json(['status' => 'error', 'message' => 'Something went wrong!!', 'exception_message' => $e]);
        }
    }

现在我希望在一个常见的地方使用删除逻辑，会有很多具有破坏功能的模型。所以我写了这个

public function destroy(Request $request)
    {
        return Crud::destroy(MyModel::class, $request);
    }

Crud Class

<?php


namespace App\Helper;


use App\Http\Controllers\Controller;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Http\Request;

class Crud extends Controller
{

    public static function destroy(Model $model,Request $request)
    {
        try {
            $output = $model::where('id', $request->get('id'))->delete();
            return response()->json([
                'status' => 'success',
                'message' => 'Deleted successfully',
                'output' => $output
            ]);
        } catch (\Exception $e) {
            return response()->json(['status' => 'error', 'message' => 'Something went wrong!!', 'exception_message' => $e]);
        }
    }

}

?>

但是当我调用destroy函数时，我收到的错误为Type error: Argument 1 passed to App\Helper\Crud::destroy() must be an instance of Illuminate\Database\Eloquent\Model, string given,

如何在函数中传递Laravel模型。

Answer 1

更改

public static function destroy(Model $model,Request $request)
{ ... }

到

public static function destroy($model,Request $request)
{ ... }

模型Model不作为实例存在。使用字符串作为模型可以起作用，但不要暗示它。

Laravel将模型类作为参数传递给函数

1 个答案: