Laravel - UploadedFile 实例的文件路径
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Laravel - file path to UploadedFile instance
提问by whonoes
I have a Laravel 4.2 API that, when creating a resource, accepts file uploads. The file is retrieved with
Input::file('file')
我有一个 Laravel 4.2 API，在创建资源时，它接受文件上传。该文件被检索
Input::file('file')
Now I want to write a script (also in Laravel) that will batch create some resources (so I can't use a HTML form that POSTs to API's endpoint). How can I translate a file path into an instance of UploadedFileso that Input::file('file')will pick it up in the API?
现在我想编写一个脚本（也在 Laravel 中）来批量创建一些资源（所以我不能使用 POST 到 API 端点的 HTML 表单）。如何将文件路径转换为 的实例，UploadedFile以便Input::file('file')在 API 中获取它？
回答by Cody Caughlan
Just construct an instance yourself. The API is:
只需自己构建一个实例。API是：
http://api.symfony.com/2.0/Symfony/Component/HttpFoundation/File/UploadedFile.html
http://api.symfony.com/2.0/Symfony/Component/HttpFoundation/File/UploadedFile.html
So you should be able to do:
所以你应该能够做到：
$file = new UploadedFile(
'/absolute/path/to/file',
'original-name.gif',
'image/gif',
1234,
null,
TRUE
);
Notice:You have to specify the 6th constructing parameter as TRUE, so the UploadedFile class knows that you're uploading the image via unit testing environment.
注意：您必须将第 6 个构造参数指定为 TRUE，以便 UploadedFile 类知道您正在通过单元测试环境上传图像。
回答by Alexandre Thebaldi
/**
* Create an UploadedFile object from absolute path
*
* @static
* @param string $path
* @param bool $public default false
* @return object(Symfony\Component\HttpFoundation\File\UploadedFile)
* @author Alexandre Thebaldi
*/
public static function pathToUploadedFile( $path, $public = false )
{
$name = File::name( $path );
$extension = File::extension( $path );
$originalName = $name . '.' . $extension;
$mimeType = File::mimeType( $path );
$size = File::size( $path );
$error = null;
$test = $public;
$object = new UploadedFile( $path, $originalName, $mimeType, $size, $error, $test );
return $object;
}
