Laravel - UploadedFile实例的文件路径

Question

我有一个Laravel 4.2 API,在创建资源时接受文件上传.使用检索文件 Input::file('file')

现在我想编写一个脚本(也在Laravel中),它将批量创建一些资源(因此我不能使用POST到API端点的HTML表单).如何将文件路径转换为实例,UploadedFile以便Input::file('file')在API中获取它？

Answer 1

只需自己构建一个实例.API是:

http://api.symfony.com/2.0/Symfony/Component/HttpFoundation/File/UploadedFile.html

所以你应该能够做到:

$file = new UploadedFile(
    '/absolute/path/to/file',
    'original-name.gif',
    'image/gif',
    1234,
    null,
    TRUE
);

注意:您必须将第6个构造参数指定为TRUE,因此UploadedFile类知道您是通过单元测试环境上传图像.

Answer 2

  /**
   * 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;
  }

Answer 3

这里是基于 Alexandre Thebaldi 答案的Laravel 8.x ：

use Illuminate\Filesystem\Filesystem;
use Illuminate\Http\UploadedFile;

/**
 * Create an UploadedFile object from absolute path 
 *
 * @param     string $path
 * @param     bool $test default true
 * @return    object(Illuminate\Http\UploadedFile)
 * 
 * Based of Alexandre Thebaldi answer here:
 * https://stackoverflow.com/a/32258317/6411540
 */
public function pathToUploadedFile( $path, $test = true ) {
    $filesystem = new Filesystem;
    
    $name = $filesystem->name( $path );
    $extension = $filesystem->extension( $path );
    $originalName = $name . '.' . $extension;
    $mimeType = $filesystem->mimeType( $path );
    $error = null;

    return new UploadedFile( $path, $originalName, $mimeType, $error, $test );
}