Efficient way to Unzip a File received through HttpPost without Unzip in Memory or File System and upload the File to Box

[Twiga Simba]

Hi,

I need guidance on a best way to accomplish the following

1. I will be receiving a zip file through an API
2. I need to upload the content of the zip to Box

I want to avoid using server memory or temporary file location to unzip and then upload the file to box. What is the best way to go about doing this. I am using .NET Core.

Here is the initial code I have where I am unzipping it to temp location

HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, uri);
                request.Headers.AcceptEncoding.Add(new StringWithQualityHeaderValue("gzip"));
                string modelString = JsonConvert.SerializeObject(model);
                request.Content = new StringContent(modelString, System.Text.Encoding.UTF8, "application/json");
                string filetemppath = string.Empty;
                string filename = string.Empty;
                HttpResponseMessage response = await client.SendAsync(request, HttpCompletionOption.ResponseHeadersRead);
                response.EnsureSuccessStatusCode();
               
                using (ZipArchive archive = new ZipArchive(await response.Content.ReadAsStreamAsync()))
                {
                    foreach (ZipArchiveEntry entry in archive.Entries)
                    {
                        using (Stream stream = entry.Open())
                        {
                            filename = entry.Name;
                            filetemppath = Path.Combine(Path.GetTempFileName());
                            using (FileStream file = new FileStream(filetemppath, FileMode.Create, FileAccess.Write))
                            {                                
                                await stream.CopyToAsync(file);                                                      
                            }
                        }
                      
                    }
                }

[Artur Jankowski]

Hi @twiga2013

Because entry.Open() returns a DeflatedStream which doesn't support Length and Position property we need to use other stream class here.
So a good solution here is to use a temp file and a FileStream like you did in your code.

Here is an example on working code:

    using (var archive = new ZipArchive(file))
    {
        foreach (ZipArchiveEntry entry in archive.Entries)
        {
            using (Stream stream = entry.Open())
            {
                var filePath = Path.GetTempFileName();

                using (var fs = new FileStream(filePath, FileMode.Open, FileAccess.ReadWrite))
                {
                    await fs.CopyToAsync(fs);
                    fs.Seek(0, SeekOrigin.Begin);

                    var fileRequest = new BoxFileRequest
                    {
                        Name = entry.Name,
                        Parent = new BoxFolderRequest { Id = "<folder_id>" }
                    };

                    var bFile = await client.FilesManager.UploadAsync(fileRequest, fs);
                    Console.WriteLine($"File {entry.Name} was uploaded successfully with id {bFile.Id}");
                }
            }
        }
    }

Hope this help!

@arjankowski

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