Question

我在serialPort.Open（）行上收到“CreateFile Failed：161”：

. . .
      MessageBox.Show(string.Format("Made it into PrintUtils.PrintBarcode()"));
using (SerialPort serialPort = new SerialPort())
{
    MessageBox.Show("Made it into using statement in PrintUtils.PrintBarcode()"); 
    serialPort.BaudRate = 19200;
    serialPort.Handshake = Handshake.XOnXOff;
    serialPort.DataBits = 8;
    serialPort.Parity = Parity.None;
    serialPort.StopBits = StopBits.One;
    serialPort.PortName = "COM1"; // Is this what it wants?
    MessageBox.Show("Made it beyond the protocol assignments in PrintUtils.PrintBarcode()"); 

    serialPort.Open(); // <-- This causes "CreateFile Failed: 161"
    MessageBox.Show("Opened the serial port in PrintUtils.PrintBarcode()"); 

    Thread.Sleep(2500); // I don't know why this is needed, or if it really is...

    // Try this first:
    serialPort.WriteLine("! 0 200 200 210 1");
    MessageBox.Show("Sent the first line in PrintUtils.PrintBarcode()"); 
    serialPort.WriteLine("TEXT 4 0 30 40 Bonjour la Monde"); //Hola el Mundo --- Hallo die Welt
    MessageBox.Show("Sent the TEXT line in PrintUtils.PrintBarcode()"); 
    serialPort.WriteLine("FORM");
    MessageBox.Show("Sent the FORM line in PrintUtils.PrintBarcode()"); 
    serialPort.WriteLine("PRINT");
    MessageBox.Show("Sent the PRINT line in PrintUtils.PrintBarcode()"); 
    // or (if WriteLine does not include a carriage return and line feed):
    //              serialPort.Write("! 0 200 200 210 1\r\n");
    //              serialPort.Write("TEXT 4 0 30 40 Bonjour la Monde\r\n"); //Hola el Mundo --- Hallo die Welt
    //              serialPort.Write("FORM\r\n");
    //              serialPort.Write("PRINT\r\n");

    serialPort.Close();
    MessageBox.Show("Closed the port in PrintUtils.PrintBarcode()"); 
}

我知道，因为我看到的最后一个“调试消息”是“超出了PrintUtils.PrintBarcode（）中的协议分配”

是因为其中一个协议错误还是格式错误？或者我是否省略了所需的协议分配？

Answer 1

错误161表示The specified path is invalid.并且您正在获取它，因为您的端口名称无效。

Windows CE要求端口名称（实际上所有驱动程序名称）后缀为“：”字符，因此您的代码应为：

serialPort.PortName = "COM1:";

当我尝试打开串口时，为什么会出现“CreateFile Failed：161”？

1 个答案: