Untitled

Delphi: open a zip archive from a stream -> extract to a stream
procedure ExtractToStream(FileName: WideString; Stream: TStream);

var
  ZipFile: TZipFile;
  DownloadedStream, DecompressionStream, MyStream: TStream;
  LocalHeader: TZipHeader;
...
ZipFile := TZipFile.Create;
try
  ZipFile.Open(DownloadedStream, zmRead);
  ZipFile.Read('myzippedfile', DecompressionStream, LocalHeader);
  try
    MyStream.CopyFrom(DecompressionStream, DecompressionStream.Size);
  finally
    DecompressionStream.Free;
  end;
finally
  ZipFile.Free;
end;

program ZipTest;

{$APPTYPE CONSOLE}

uses
  System.SysUtils,
  System.Classes,
  System.Zip;

procedure ExtractToFile(
  const ZipFileName: string;
  const ZippedFileIndex: Integer;
  const ExtractedFileName: string
);
var
  ZipFile: TZipFile;
  DownloadedStream, DecompressionStream, OutputStream: TStream;
  LocalHeader: TZipHeader;
begin
  DownloadedStream := TFileStream.Create(ZipFileName, fmOpenRead);
  try
    ZipFile := TZipFile.Create;
    try
      ZipFile.Open(DownloadedStream, zmRead);
      ZipFile.Read(ZippedFileIndex, DecompressionStream, LocalHeader);
      try
        OutputStream := TFileStream.Create(ExtractedFileName, fmCreate);
        try
          OutputStream.CopyFrom(DecompressionStream, DecompressionStream.Size);
        finally
          OutputStream.Free;
        end;
      finally
        DecompressionStream.Free;
      end;
    finally
      ZipFile.Free;
    end;
  finally
    DownloadedStream.Free;
  end;
end;

begin
  try
    ExtractToFile('C:desktoptest.zip', 0, 'C:desktopout.txt');
  except
    on E: Exception do
      Writeln(E.ClassName, ': ', E.Message);
  end;
end.