获取上传文件的原始名称 [英] Get Uploaded File's Original Name

问题描述

$request = $event->getRequest();
print_r($request->files);die;

给我

Symfony\Component\HttpFoundation\FileBag Object
(
    [parameters:protected] => Array
        (
            [files] => Array
                (
                    [0] => Symfony\Component\HttpFoundation\File\UploadedFile Object
                        (
                            [test:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 
                            [originalName:Symfony\Component\HttpFoundation\File\UploadedFile:private] => Chrysanthemum.jpg
                            [mimeType:Symfony\Component\HttpFoundation\File\UploadedFile:private] => image/jpeg
                            [size:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 879394
                            [error:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 0
                            [pathName:SplFileInfo:private] => /tmp/phpmUl54W
                            [fileName:SplFileInfo:private] => phpmUl54W
                        )

                )

        )

)

我试图在不使用循环的情况下获取原始名称"即Chrysanthemum.jpg"的值，但我似乎找不到正确的语法

I'm trying to get at the value for 'originalname' i.e "Chrysanthemum.jpg" without resorting to a loop, but I can't seem to find the right syntax

使用 1UP 文件上传器，但我认为这并不重要

Using the 1UP file uploader, but I dont think that's important

推荐答案

这对我有用，我猜 OneUp Class 处理它有点不同

This wound up working for me, I guess the OneUp Class handles it a bit differently

use Oneup\UploaderBundle\Event\PostPersistEvent;
class UploadListener
{
     public function onUpload(PostPersistEvent $event)
    {

        $request = $event->getRequest();
        $original_filename = $request->files->get('blueimp')->getClientOriginalName();
     }
}

相关前端

<input id="fileupload" type="file" name="blueimp" data-url="{{ oneup_uploader_endpoint('images') }}" multiple />

这篇关于获取上传文件的原始名称的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！