Question

我在上传现有文件夹中的照片时遇到问题。当我上传新照片例如（1）.jpg时，我会用空格分割它，获取前例名称，创建一个具有该名称的文件夹并将图片上传到该文件夹中。所有罚款和花花公子，但当我上传另一张照片forexample（2）.jpg进入forexample文件夹laravel给我错误无法写入＆＃34; ...... / forexample /＆＃34; 。目录

代码看起来像这样。它在控制器内。谢谢你事先偷看。

if(Input::file('poze')[0] != null)
{
    //make folder name
    $foldername = explode(' ',Input::file('poze')[0]->getClientOriginalName())[0];

    //make folder path
    $path = substr(storage_path(),0,27).'pozeproduse/';

    if(File::exists($path.$foldername) == false){

        //create folder
        File::makeDirectory($path.$foldername,0777);

        Input::file('poze')[0]->move($path.$foldername.'/',Input::file('poze')[0]->getClientOriginalName());
        $poze = new Poze();
        $poze->poza = substr($path.$foldername.'/'.Input::file('poze')[0]->getClientOriginalName(),26,64);
        $poze->produs_id = $this->request->input('idp');
        $poze->save();


    }else{
        //dd($path.$foldername);
        //foreach(scandir($path.$foldername) as $d){
        //  if($d != Input::file('poze')[0]){

            $obj= new \stdClass();
            $obj->poza = Input::file('poze')[0];

            $obj->poza->move($path.$foldername.'/',Input::file('poze')[0]->getClientOriginalName());
            //dd('break');

            $poze = new Poze();
            $poze->poza = substr($path.$foldername.'/'.Input::file('poze')[0]->getClientOriginalName(),26,64);
            $poze->produs_id = $this->request->input('idp');
            $poze->save();
        //  }
        //}
    }
}

无法写入...目录Laravel

0 个答案: