asp.net - C# Prompt for open/save zip in ASP receiving error -

let me start off saying i'm sure that's quite simple, unfortunately can't seem figure out. here's problem, query database, return need, zip up, , prompt user save. when prompt appears, appears system.io.compression.ziparchive.zip file name. when attempt open this, says file invalid. appreciated!!

if (e.commandname == "downloadattachment")     {         e.canceled = true;         // create zip , send client.         //response.write(@"<script language='javascript'>alert('details saved successfully')</script>");         var item = e.item grideditableitem;         fileid = (int)item.getdatakeyvalue("unique");         filedata[] allrecords = null;         using (             sqlconnection conn =                 new sqlconnection(configurationmanager.connectionstrings["ptcdbmodelentities"].connectionstring))         {             using (                 sqlcommand comm = new sqlcommand("select unique1, binarydata, filename ptcdbtracker.dbo.caffiletable unique1=@fileid , filename not null", conn))             {                 comm.parameters.add(new sqlparameter("@fileid", fileid));                 conn.open();                 using (var reader = comm.executereader())                 {                     var list = new list<filedata>();                     while (reader.read())                     {                         list.add(new filedata { unique1 = reader.getint32(0) });                         long len = reader.getbytes(1, 0, null, 0, 0);                         byte[] buffer = new byte[len];                         list.add(new filedata { binarydata = (byte)reader.getbytes(1, 0, buffer, 0, (int)len), filename = reader.getstring(2) });                         allrecords = list.toarray();                     }                 }                 conn.close();             }         }          using (var compressedfilestream = new memorystream())         {             //create archive , store stream in memory.              using (var ziparchive = new ziparchive(compressedfilestream, ziparchivemode.update, false))             {                 if (allrecords != null)                 {                     foreach (var record in allrecords)                     {                         //create zip entry each attachment                         if (record.filename != null)                         {                              var zipentry = ziparchive.createentry(record.filename);                              //get stream of attachment                             using (var originalfilestream = new memorystream(record.binarydata))                             {                                 using (var zipentrystream = zipentry.open())                                 {                                     //copy attachment stream zip entry stream                                     originalfilestream.copyto(zipentrystream);                                 }                             }                         }                     }                 }                 response.clearcontent();                 response.clearheaders();                 response.binarywrite(compressedfilestream.toarray());                 response.appendheader("content-disposition", "attachment; filename=result.zip");                 response.flush();                 response.close();                 ziparchive.dispose();                 //how prompt open or save?             }         } 

edit: included sami's suggestion. zip folder saves , opens, not have content.

you send correct content-disposition header, filename append object, platform calls tostring() on , default output class name. want send either random filename, or static one.

response.appendheader("content-disposition", "attachment; filename=result.zip"); 

then sending of file. use memorystream save zip archive , don't send it. 1 way be

response.binarywrite(compressedfilestream.toarray()); response.flush(); response.close();