我有刺蛾,輸入文件形式的極品具體名稱等Laravel - 文件和文件夾存儲功能
,如果我把POR例如,在形式,我需要:
slug = hello-how-are-you
file = header.png
file2 = home.png
Laravel
創建我的文件夾中像THI存儲s:/projects/hello-how-are-you
與header.png
和home.png
。
我嘗試,但不能做到這一點..我的東西嘗試這樣的:
public function storeProject(Request $request)
{
$project = new Project();
$project->slug = $request->input("slug");
$namefolder = $project->slug;
$project->position = $request->input("position");
$project->public = $request->input("public");
$header = $request->file('pathheader');
$home = $request->file('pathhome');
$project->pathheader = $header;
$project->pathhome = $home;
\Storage::disk('projects')->put('header.png', \File::get($header));
$project->save();
}
在文件系統中,我有這樣的:
'disks' => [
'local' => [
'driver' => 'local',
'root' => storage_path('app'),
],
'public' => [
'driver' => 'local',
'root' => storage_path('app/public'),
'url' => env('APP_URL').'/storage',
'visibility' => 'public',
],
'projects' => [
'driver' => 'local',
'root' => storage_path() . '/projects',
],
's3' => [
'driver' => 's3',
'key' => env('AWS_KEY'),
'secret' => env('AWS_SECRET'),
'region' => env('AWS_REGION'),
'bucket' => env('AWS_BUCKET'),
],
],
解決了! 感謝@ Dees040 如果要粉碎的文件的名稱和做沒有一個foreach,你可以做這樣的:
public function storeProject(Request $request)
{
$project = new Project();
$project->slug = $request->input("slug");
$project->position = $request->input("position");
$project->public = $request->input("public");
$project->pathheader = $request->file('pathheader');
$project->pathhome = $request->file('pathhome');
\Storage::disk('projects')->makeDirectory($project->slug);
\Storage::disk('projects')->putFileAs($project->slug,$project->pathheader,'header.png');
\Storage::disk('projects')->putFileAs($project->slug,$project->pathhome,'home.png');
$project->save();
}
不'\存儲::磁盤( '項目') - >把($項目 - >蛞蝓。「/ header.png」)'工作? – apokryfos
@apokryfos沒有伴侶,如果我這樣做的文件名將是你好,你是怎麼樣,youheader.png hahah –
所以有什麼東西吞噬斜線? – apokryfos