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By: a guest on May 4th, 2012  |  syntax: None  |  size: 1.38 KB  |  hits: 19  |  expires: Never
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  1. System.Diagnostics.Process.Start() doesn't work in the web server
  2. MySqlDataReader Reader = null;
  3. connection.Open();
  4. MySqlCommand command = new MySqlCommand("Select Image, File_Type, File_Name from table where ImageID = " + ImageID, connection);
  5. Reader = command.ExecuteReader();
  6.  
  7. if (Reader.Read())
  8. {
  9.     byte[] buffer = (byte[])Reader["Image"];
  10.     System.IO.MemoryStream stream1 = new System.IO.MemoryStream(buffer, true);
  11.     stream1.Write(buffer, 0, buffer.Length);
  12.  
  13.     String fileName = Reader["File_Name"].ToString();
  14.     String dirName = "C:\thefolder\";
  15.     if (!Directory.Exists(dirName))
  16.     {
  17.         // if not then create
  18.         Directory.CreateDirectory(dirName);
  19.     }
  20.     if (File.Exists(dirName+fileName))
  21.         File.Delete(dirName + fileName);
  22.     Directory.CreateDirectory(Path.GetDirectoryName(Reader["File_Name"].ToString()));
  23.  
  24.     using (Stream file = File.Create(dirName + fileName))
  25.     {
  26.         file.Write(buffer, 0, buffer.Length);
  27.     }
  28.     Process process = new Process();
  29.     process.StartInfo.FileName = "AcroRd32.exe";
  30.     process.Start();                    
  31. }
  32.        
  33. //Process process = new Process();
  34. //process.StartInfo.FileName = "AcroRd32.exe";
  35. //process.Start();
  36. Response.ClearHeaders();
  37. Response.ContentType = "application/pdf";
  38. Response.Clear();
  39. Response.AppendHeader("Content-Disposition", "attachment");
  40. Response.TransmitFile(dirName + fileName);
  41. Response.End();