将图像复制到存储文件夹
将图像复制到存储文件夹
提问于 2018-12-26 20:53:02
我正在尝试使用以下代码将图像复制到storage/app/uploads文件夹：
public function save(Request $request)
{
$arrayName = array();
echo "<pre>";
print_r($request->all());
echo "</pre>";
$image = $request->file('image');
$name = $imagem->getClientOriginalName();
$ext = $imagem->getClientOriginalExtension();
$newName = str_replace(' ','_', $imagem->getClientOriginalName());
$destinationPath = 'uploads';
//$path = $imagem->store($newName);
$path = $imagem->storeAs($destinationPath, $newName);
echo "<pre>";
print_r($path);
echo "</pre>";
}以上代码的结果是：
Array
(
[id] =>
[description] => description
[title] => title
[title_small] => title_small
[text] => text
[image] => Illuminate\Http\UploadedFile Object
(
[test:Symfony\Component\HttpFoundation\File\UploadedFile:private] =>
[originalName:Symfony\Component\HttpFoundation\File\UploadedFile:private] => manaus1.png
[mimeType:Symfony\Component\HttpFoundation\File\UploadedFile:private] => application/octet-stream
[error:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 1
[hashName:protected] =>
[pathName:SplFileInfo:private] =>
[fileName:SplFileInfo:private] =>
)
[published] => S
)但是，当我打开保存的内容时，它会打开一个带有原始扩展名的文本文档。
我要用正确的方式保存。
回答 2
Stack Overflow用户
回答已采纳
发布于 2018-12-26 21:15:09
我在用这段代码保存
if(Input::hasFile('imagen')) {
$time = Carbon::now()->format('Y-m-d');
$image = $request->file('imagen');
$extension = $image->getClientOriginalExtension();
$name = $image->getClientOriginalName();
$fileName = $time."-".$name;
$image->move(storage_path(),$fileName);
}请试试这个，让我知道它的工作原理。:)
Stack Overflow用户
发布于 2018-12-26 21:29:24
我建议你用这个：
// SAME VALUES FROM THE FIRST ANSWER BUT WITH ANOTHER UPLOAD SYSTEM
if(Input::hasFile('imagen')) {
$time = Carbon::now()->format('Y-m-d');
$image = $request->file('imagen');
$extension = $image->getClientOriginalExtension();
$name = $image->getClientOriginalName();
$fileName = $time."-".$name;
// UPLOAD AND RUN YOUR SQL CODE IF YOU WANT ...
$path = \Storage::putFile('your_path', $image);
# LARAVEL WILL GENRATE A UNIQUE FILE NAME ;)
return dd($path);
// OR RETURN JSON RESPONSE IF YOU USE AJAX ;)
return response()->json([
'title' => $fileName,
'path' => $path,
'status' => 'success'
]);
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/53937049
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