如何上傳輸入文件在不同的文件夾,我需要使用開關的情況下做出不同的文件夾中的每個輸入文件,這樣做的代碼,但是當我執行它沒有發生,我想知道在哪裏的問題laravel使用開關的情況下上傳文件
我看來
{!! 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CONTROLER
$model = new Files($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;
$model->save();
}
break;
}
如果DD($模型)它返回什麼?此外,當你做一個開關就會使情況相匹配的第一個後停止,所以如果你想要做多,你需要做的輸入一個foreach和運行它們通過交換機 – rchatburn
你能顯示代碼爲模型? –
這看起來像可疑的https://stackoverflow.com/questions/45075572/laravel-upload-files-in-many-inputs/45075789#45075789 –