为什么我得到“;创建文件失败:161”;当我试图打开串行端口时
本文关键字:串行端口 创建 文件 为什么 失败 | 更新日期: 2023-09-27 17:58:19
我在serialPort.Open()行上得到"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()");
}
我知道这一点,因为我看到的最后一条"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不正确?或者我遗漏了所需的协议分配?
错误161表示The specified path is invalid.
,您得到它是因为您的端口名无效。
Windows CE要求端口名(实际上是所有驱动程序名)以":"字符作为后缀,因此您的代码应该是:
serialPort.PortName = "COM1:";