如何在没有表单的情况下获取文件上传 [英] How to get file upload without a form

问题描述

我的控制器接收到发布数据.它不是来自 Symfony 生成的表单，而是来自使用 FormData 的 AngularJS 自定义表单.

my controller receives post data. It's not from a Symfony generated form, but from an AngularJS custom form using FormData.

在请求参数包中正常接收正常参数，但是如何获取上传的文件?我是否需要手动执行与表单的 handleRequest 相同的操作?

The normal parameters are received normally in the request parameter bag, but how do I get the file that is uploaded? Do I need to manually do the same that the form's handleRequest does?

我的文档与 cookbook 文章相同.

所以我需要从 post 请求中获取 UploadFile 以在文档实体上调用此方法:

So I need to get an UploadFile from the post request to call this method on the document entity:

公共函数 setFile(UploadedFile $file = null)

如何在 Symfony 2 中手动接收上传(不以任何方式使用表单生成器)?

How do I receive a upload manually in Symfony 2 (without using the form builder in any way)?

推荐答案

请求 有一个FileBag，类似于ParameterBag

The Request has a FileBag, similar to the ParameterBag

所以我可以很容易地得到指定的文件:

So I could get the file specified easily with:

$data = $this->getRequest()->request->all();
$file = $this->getRequest()->files->get('file');

并按原样使用食谱中的文档:

and use the document as is from the cookbook:

$document = new Document();
$document->setFile($file);
$lead->setDocument($document);

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