laravel使用开关盒上载文件

如何上传不同文件夹中的输入文件我需要使用开关盒使每个输入文件位于不同的文件夹中，并执行了此代码，但当我执行时，什么都没有发生，我想知道问题出在哪里

我的看法

{!! Form::file('file1', null,['class'=>'form-control']) !!}
{!! Form::file('file2', null,['class'=>'form-control']) !!}
{!! Form::file('file3', null,['class'=>'form-control']) !!}
{!! Form::file('file4', null,['class'=>'form-control']) !!}

我的控制者

$model = new Files($request->all());
        switch ($model) {
            case "file1":
                if ($request->hasFile('file1')) {
                    $file = $request->file('file1');
                    $destinationPath = public_path() . '/file1';
                    $filename = $file->getClientOriginalName();
                    $file->move($destinationPath, $filename);
                    $request['file1'] = $filename;
                    $model -> file1 = $filename;
                    $model->save();
                }
                break;
            case "file2":
                if ($request->hasFile('file2')) {
                    $file = $request->file('file2');
                    $destinationPath = public_path() . '/file2';
                    $filename = $file->getClientOriginalName();
                    $file->move($destinationPath, $filename);
                    $request['file2'] = $filename;
                    $model->file2 = $filename;
                    $model->save();
                }
                break;
                case "file3":
                if ($request->hasFile('file3')) {
                    $file = $request->file('file3');
                    $destinationPath = public_path() . '/file3';
                    $filename = $file->getClientOriginalName();
                    $file->move($destinationPath, $filename);
                    $request['file3'] = $filename;
                    $model->file3 = $filename;
                    $model->save();
                }
                    break;
            case "file4":
                if ($request->hasFile('file4')) {
                    $file = $request->file('file4');
                    $destinationPath = public_path() . '/file4';
                    $filename = $file->getClientOriginalName();
                    $file->move($destinationPath, $filename);
                    $request['file4'] = $filename;
                    $model->file4 = $filename;
                    $model->save();
                }
                break;
        }

---

嗯，是的，我做过类似的事情，但我用

foreach

循环，然后用

if

这是我的示例代码，我希望它能帮助：-D

$requests = $request->all();
$model =  new File;
foreach ($requests as $key => $val) {
   if ($key == 'file1' && !empty($val)) {

       $destinationPath = public_path() . '/file1';
       $filename = $val->getClientOriginalName();
       $val->move($destinationPath, $filename);
       $model -> file1 = $filename;
       $model->save();
   }
   if ($key == 'file2' && !empty($val)) {

       $destinationPath = public_path() . '/file2';
       $filename = $val->getClientOriginalName();
       $val->move($destinationPath, $filename);
       $model -> file1 = $filename;
       $model->save();
   }
   // And do it again as much as you need :D
}

---

给你，我希望它能帮助你朋友：D

如果你dd（$model），它会返回什么？此外，当您执行切换时，它将在第一个匹配案例的切换后停止，因此如果您要执行多个切换，则需要在输入中执行foreach并通过切换运行它们，您是否可以显示模型的代码？这看起来可疑地像否，它不像foldersmu model code protected$table中的链接此文件='files'；受保护的$fillable=['file1'，'file2'，'file3'，'file4']；A很高兴能帮助你：D