从 web api 返回 zip 文件

Question

I have built an asp net web api. I need to return a zipfile, as a result of some inner logic. I'm using this code and it works, but the resulting zip file, when unzipped manually, gave me this error "There are data after the end of the payload"

 using (ZipFile zip = new ZipFile())
 {
     ...
     zip.Save(di.FullName + "\\" + "Update.zip");
 }


 string path = Path.Combine(Properties.Settings.Default.PathDisposizioniHTML, "Update.zip");

 var response = new HttpResponseMessage(HttpStatusCode.OK);
 var stream = new System.IO.FileStream(path, System.IO.FileMode.Open);
 response.Content = new StreamContent(stream);
 response.Content.Headers.ContentType = new System.Net.Http.Headers.MediaTypeHeaderValue("application/octet-stream");

This is how i receive the data in a .net console application:

using (Stream output = File.OpenWrite(@"C:\prova\MyFile.zip"))
    using (Stream input = httpResponse.GetResponseStream())
    {
        input.CopyTo(output);
    }

Answer 1

If you already have the zip file on your system, you shouldn't need to do anything special before sending it as a response.

This should work:

string filePath = @"C:\myfolder\myfile.zip";

return File(filePath, "application/zip");

If you're making the file on the fly, ie getting other files and programatically putting them into a zip file for the user, the following should work:

public IActionResult GetZipFile(){

   //location of the file you want to compress
   string filePath = @"C:\myfolder\myfile.ext";

   //name of the zip file you will be creating
   string zipFileName = "zipFile.zip";

   byte[] result;


   using (MemoryStream zipArchiveMemoryStream = new MemoryStream())
   {
       using (ZipArchive zipArchive = new ZipArchive(zipArchiveMemoryStream, ZipArchiveMode.Create, true))
       {
           ZipArchiveEntry zipEntry = zipArchive.CreateEntry(zipFileName);
           using (Stream entryStream = zipEntry.Open())
           {
               using (MemoryStream tmpMemory = new MemoryStream(System.IO.File.ReadAllBytes(filePath)))
               {
                    tmpMemory.CopyTo(entryStream);
                };
           }
       }

       zipArchiveMemoryStream.Seek(0, SeekOrigin.Begin);
       result = zipArchiveMemoryStream.ToArray();
   }

   return File(result, "application/zip", zipFileName);

}

This is taken from a recent ASP.NET project of my own.