Laravel-UploadedFile实例的文件路径 [英] Laravel - file path to UploadedFile instance

问题描述

我有一个Laravel 4.2 API，当创建资源时，它接受文件上传.使用以下方式检索文件 Input::file('file')

I have a Laravel 4.2 API that, when creating a resource, accepts file uploads. The file is retrieved with Input::file('file')

现在，我想编写一个脚本(也在Laravel中)，该脚本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UploadedFile的实例，以便Input::file('file')将其从API中提取?

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 UploadedFile so that Input::file('file') will pick it up in the API?

推荐答案

只需自己构造一个实例.该API是:

Just construct an instance yourself. The API is:

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

因此您应该可以:

$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类知道您正在通过单元测试环境上传图像.

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.

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