IRemoteStreamContent - Download Large Files #9354

13 Answer(s)

• 

  0

  EngincanV created 22 days ago

  Support Team .NET Developer

  Hi, ABP’s RemoteStreamContent is great for lightweight files or when you want to abstract things but it’s not designed for very large files. It attempts to copy the stream into an in-memory buffer (MemoryStream), which crashes for files >2GB or causes heavy memory usage.

  So, for large files, instead, you can use FileStream (see https://stackoverflow.com/a/39890811/10477283) and FileStreamResult.

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• 

  0

  kkmy created 22 days ago

  Is there an example for this? Because I have written it like the following:

  public async Task< **FileStreamResult** > ExportAsync()
  {
      ...
      
      return new FileStreamResult(zipFileStream, FileTypes.Zip.ContentType);
  }
  

  However, proxy generation is failed:

  

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• 

  0

  maliming created 22 days ago

  Support Team Fullstack Developer

  hi

  I am trying to send the file to the front end so that the user can download it.

  Please share your full app service code.

  Thanks.

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• 

  0

  kkmy created 22 days ago

  public async Task< FileStreamResult > ExportAsync(ExportDto input)
  {
      var contents = new Dictionary<string, byte[]>();
      var discardedContents = new Dictionary<string, byte[]>();
  
      FillMuyCompressionItems(ref contents, ref discardedContents);
      FillMlCompressionItems(ref contents, ref discardedContents);
      FillAkbCompressionItems(ref contents, ref discardedContents);
      FillCkCompressionItems(ref contents, ref discardedContents);
  
      var zipFileName = "EXPORT.zip";
  
      var mainZipFileStream = new FileStream(Path.Join(Path.GetTempPath(), zipFileName), FileMode.Create, FileAccess.ReadWrite);
  
      using (var zipArchive = new ZipArchive(mainZipFileStream, ZipArchiveMode.Create, true))
      {        
          for (var i = 0; i < contents.Count; i++)
          {
              var item = contents.ElementAt(i);
          
              var originalFileStream = new FileStream(Path.Join(Path.GetTempPath(), item.Key), FileMode.Create, FileAccess.ReadWrite);
              await originalFileStream.WriteAsync(item.Value);
              originalFileStream.Seek(0, SeekOrigin.Begin);
          
              var zipEntry = zipArchive.CreateEntry(item.Key, CompressionLevel.Fastest);
  
              await using(var zipEntryStream = zipEntry.Open())
              await using (originalFileStream)
              {
                  await originalFileStream.CopyToAsync(zipEntryStream);
              }
          }
      }
  
      mainZipFileStream.Seek(0, SeekOrigin.Begin);
  
      return new FileStreamResult(mainZipFileStream, FileTypes.Zip.ContentType);
  }
  

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• 

  0

  maliming created 22 days ago

  Support Team Fullstack Developer

  hi

  Can you also share the code of IRemoteStreamContent version? Thanks

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• 

  0

  kkmy created 22 days ago

  public async Task< IRemoteStreamContent > ExportAsync(ExportDto input)
  {
      var contents = new Dictionary<string, byte[]>();
      var discardedContents = new Dictionary<string, byte[]>();
  
      FillMuyCompressionItems(ref contents, ref discardedContents);
      FillMlCompressionItems(ref contents, ref discardedContents);
      FillAkbCompressionItems(ref contents, ref discardedContents);
      FillCkCompressionItems(ref contents, ref discardedContents);
  
      var zipFileName = "EXPORT.zip";
  
      var mainZipFileStream = new FileStream(Path.Join(Path.GetTempPath(), zipFileName), FileMode.Create, FileAccess.ReadWrite);
  
      using (var zipArchive = new ZipArchive(mainZipFileStream, ZipArchiveMode.Create, true))
      {
          for (var i = 0; i < contents.Count; i++)
          {
              var item = contents.ElementAt(i);
              
              using var originalFileStream = new MemoryStream(item.Value);
  
              var zipEntry = zipArchive.CreateEntry(item.Key, CompressionLevel.Fastest);
  
              await using var zipEntryStream = zipEntry.Open();
  
              await originalFileStream.CopyToAsync(zipEntryStream);
          }
      }
  
      mainZipFileStream.Seek(0, SeekOrigin.Begin);
  
      return new RemoteStreamContent(mainZipFileStream, zipFileName, FileTypes.Zip.ContentType);
  }
  

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• 

  0

  maliming created 22 days ago

  Support Team Fullstack Developer

  Thanks. Do you have app.UseAbpStudioLink(); in your IPSHttpApiHostModule.cs?

  If so, Please remove it and use IRemoteStreamContent again.

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• 

  0

  kkmy created 22 days ago

  It worked perfectly, thank you so much!

  Though, I really wonder what is the corelation between AbpStudioLink and IRemoteStreamContent if it is not a secret? :) What makes it does not work?

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• 

  0

  maliming created 22 days ago

  Support Team Fullstack Developer

  hi

  AbpStudioLink will use MemoryStream; We will fix it in the next version.

  Thanks.

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• 

  0

  kkmy created 22 days ago

  Oh I see!

  Does this mean that it is better to add app.UseAbpStudioLink(); again after the next version? Or let it be removed like this?

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• 

  0

  maliming created 22 days ago

  Support Team Fullstack Developer

  hi

  You can use it in the next version. remove it now. ; )

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• 

  0

  kkmy created 22 days ago

  Got it!

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• 

  0

  maliming created 22 days ago

  Support Team Fullstack Developer

  Great

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