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HomeHomeDevelopment and...Development and...Building ExtensionsBuilding ExtensionsModulesModulesZipping files from a "database" folderZipping files from a "database" folder
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7/25/2012 5:32 AM
 

Hi,

I am developing a module that allows a user to download files from a given folder packed in a Zip file. This works fine for Standard and Secure folders, but I don't have any idea how to solve this for Database folders. I am using DotNetZip to realize this, and the code to create the Zip package so far is:

            using (ZipFile zip = new ZipFile())
            {
               for (int i = 0; i < files.Count; i++)
               {
                  switch (folder.StorageLocation)
                  {
                     case 0:  // Normal
                        zip.AddFile(files[i].PhysicalPath);
                        break;
                     case 1:  // Secure
                        ZipEntry newFile = zip.AddFile(String.Format("{0}.resources", files[i].PhysicalPath));
                        newFile.FileName = files[i].FileName;
                        break;
                     default:
                        // Database folders - how????
                        break;
                  }
               }
               zip.Save(Response.OutputStream);
            }

Is there someone who is able to point me in the right direction?

Thanks!
Michael


Michael Tobisch
DNN★MVP

dnn-Connect.org - The most vibrant community around the DNN-platform
 
New Post
7/25/2012 10:24 AM
 

OK, got it...

            using (ZipFile zip = new ZipFile())
            {
               for (int i = 0; i < files.Count; i++)
               {
                  switch (folder.StorageLocation)
                  {
                     case 0:  // Normal
                        zip.AddFile(files[i].PhysicalPath);
                        break;
                     case 1:  // Secure
                        ZipEntry newFile = zip.AddFile(String.Format("{0}.resources", files[i].PhysicalPath));
                        newFile.FileName = files[i].FileName;
                        break;
                     case 2:  // Database
                        Stream fileContent = FileManager.Instance.GetFileContent(files[i]);
                        zip.AddEntry(files[i].FileName, fileContent);
                        break;
                     default:
                        break;
                  }
               }
               zip.Save(Response.OutputStream);
            }
I didn't expect it's that easy... Best wishes Michael

Michael Tobisch
DNN★MVP

dnn-Connect.org - The most vibrant community around the DNN-platform
 
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