问题描述
我有一个 Laravel 4.2 API，它在创建资源时接受文件上传.该文件是用输入::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)，它将批量创建一些资源(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 UploadedFile 的实例，以便 Input::file('file') 将在 API 中获取它?
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
自己构造一个实例即可.API 是:
Just construct an instance yourself. The API is:
http://api.symfony.com/2.0/Symfony/Component/HttpFoundation/File/UploadedFile.html
所以你应该可以做到:
$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，以便 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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