C＃IFormFile作为ZipFile [英] C# IFormFile as ZipFile

问题描述

我有一个REST API端点，该端点在.Net Core 1.1上接收zip文件。我从这样的请求中获取IFormFile

I have a REST API endpoint which receives zip file on .Net Core 1.1. I'm getting IFormFile from request like this

var zipFile = HttpContext.Request.Form.Files.FirstOrDefault();

然后我需要将其传递给.Net Standard 1.5中的服务方法，其中不支持IFormFile 。

And then I need to pass it to service method from .Net Standard 1.5, where IFormFile is not supported.

所以问题是：如何将IFormFile转换为ZipFile或Standard 1.5支持的其他类型，或者有一些更合适的操作方式压缩文件？
谢谢！

So the question is: how can I convert IFormFile to ZipFile or to some other type which is supported in Standard 1.5, or maybe there is some more proper way to operate with zip files? Thanks!

推荐答案

IFormFile 只是一个包装收到的文件。您仍然应该阅读实际文件。例如，您可以将文件流读取为字节数组，并将其传递给服务：

IFormFile is just a wrapper for the received file. You should still read the actual file do something about it. For example, you could read the file stream into a byte array and pass that to the service:

byte[] fileData;
using (var stream = new MemoryStream((int)file.Length))
{
    file.CopyTo(stream);
    fileData = stream.ToArray();
}

或者您也可以将流复制到文件系统中的物理文件中。

Or you could copy the stream into a physical file in the file system instead.

但这基本上取决于您实际要对上传的文件进行什么操作，因此您应该从该方向开始并转换 IFormFile 放入所需的内容。

But it basically depends on what you actually want to do with the uploaded file, so you should start from that direction and the convert the IFormFile into the thing you need.

如果要以ZIP格式打开文件并解压缩您可以尝试使用 ZipArchive 构造函数，需要一个流。像这样的东西：

If you want to open the file as a ZIP and extract something from it, you could try the ZipArchive constructor that takes a stream. Something like this:

using (var stream = file.OpenReadStream())
using (var archive = new ZipArchive(stream))
{
    var innerFile = archive.GetEntry("foo.txt");
    // do something with the inner file
}

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