title: How to verify if Zebra printer successfully printed using ZPL and C# (or be able to detect errors)?
body: <h2>Problem</h2>
  
  <p>Is there a simple ZPL code or way to get an error message back from a Zebra printer to determine if the labels did not print successfully or that there was some kind of error? </p>
  
  <h2>Progress</h2>
  
  <p>Here is a nice function I built to send a printer job to the zebra printer: </p>
  
  <pre><code>public static void SendToPrinter(string zplString, string ipAddress = "127.0.0.1", int port = 1337)
          {
              // Open connection
              TcpClient tcpClient = new TcpClient();
              tcpClient.Connect(ipAddress, port);
  
              // Write ZPL String to connection
              using (System.IO.StreamWriter writer = new System.IO.StreamWriter(tcpClient.GetStream()))
              {
                  writer.Write(zplString);
                  writer.Flush();
                  writer.Close();
              }
              // Close Connection
              tcpClient.Close();
          }
  </code></pre>
  
  <p>There is a lot of magic happening in the zplString, but basically it consists of the ZPL code we all have come to love. The problem in my approach is that it seems this is a rather one way ticket to the printer. A lot of work went into the above and I am hoping we can somehow modify it to listen for a response if I somehow had the appropriate ZPL code to listen to a response? </p>
  
  <p>I simply have not seen any literature or forum discuss how to receive a response back from a zebra printer and determine if it was successful? </p>
  
  <h2>Issues</h2>
  
  <p>Ideally, I would like a way to understand the printer using ZPL wrapped inside C# and .NET if the printer succeeded or failed in some way. Otherwise, I might have to manually query the user "did it print?". This is not ideal, but I have not yet found anything in my manual that indicated an easy way to detect that there was an error in the print job using ZPL? </p>
  
  <p>Thanks for your patience, assistance, and for reading this question. </p>