CODEWITHSUNDEEP
c#filestreamzip
B5-NDDT
B5-NDDT

Reputation: 187

Get Filestream from ZipArchive

I have a function that requires a Filestream as input.

I want to hand several Files to that function which I get from uploaded zip-Files. Is it possible to create the Filestream without extracting the file to a temporary folder?

I imagin something like this:

string path = @"C:\somepathtomyzip";
string filepath = "nameofimagefile"
using (ZipArchive archive = ZipFile.OpenRead(path))
{
    ZipArchiveEntry entry = archive.GetEntry(file_path);

    //generate Filestream from entry

    myFunction(filestreamIneed);
}

Upvotes: 0

Views: 2725

Answers (1)

Alex
Alex

Reputation: 21766

You can use ZipArchiveEntry.Open() and copy the output from the returned Stream instance to a FileStream instance:

using (ZipArchive archive = ZipFile.OpenRead(path))
{
    ZipArchiveEntry entry = archive.GetEntry(file_path);

     var memoryStream = return entry.Open();

     using (var fileStream = new FileStream(fileName, FileMode.CreateNew, FileAccess.ReadWrite))
     {
         memoryStream.CopyTo(fileStream); // fileStream is not populated
     }
}

Upvotes: 2

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