title: FileSystemWatcher - event not firing the second time
body: <p>I have an application that launches other applications, and then waits for them to create a specific data file (it watches one application at a time). Each time an application is launch it watches a specific directory for a specific file to be created. I am using the FileSystemWatcher to do this (set it to the directory, then filter for the correct file name). This works great the first time (always), but the second application launched never fires the event. The only way it seems to fire the event is if I place a break-point in the event handler, or if I have a Thread.Sleep command in the event handler. This seems very strange to me...is there some race condition that I'm not aware of? Here is the code. Notice I have a Thread.Sleep(500). With this line the code works every time. Without it will fail. I'm really not comfortable relying on a Sleep command. I'm not sure what condition will cause that not to work as well.</p>
  
  <pre><code>    public static void watchFiles(string path)
      {
          FileSystemWatcher watcher = new FileSystemWatcher();
          watcher.Path = path;
          watcher.Created += new FileSystemEventHandler(watcher_Handler);
          watcher.EnableRaisingEvents = true;
     }
  
      public static void watcher_Handler(object sender, FileSystemEventArgs e)
      {
          //Hack - the sleep allows the second and third application to be caught by this event
          Thread.Sleep(500);
  
          switch (e.ChangeType.ToString())
          {
              case "Changed":
                  break;
              case "Deleted":
                  break;
              case "Created":
                  if (e.Name == "log.dat")
                  {
                      parseDataFile();
                      moveHTMLtoLMS();
  
                  }
                  break;
              default:
                  break;
          }
      }
  </code></pre>
  
  <p>Anyone know why I need to have that Sleep (or break-point) to get the code to work a second time?</p>