我正在开发需要与COM端口交互的程序.
通过学习这个问答:.NET SerialPort DataReceived事件没有触发,我就像这样编写代码.
namespace ConsoleApplication1
{
class Program
{
static SerialPort ComPort;
public static void OnSerialDataReceived(object sender, SerialDataReceivedEventArgs args)
{
string data = ComPort.ReadExisting();
Console.Write(data.Replace("\r", "\n"));
}
static void Main(string[] args)
{
string port = "COM4";
int baud = 9600;
if (args.Length >= 1)
{
port = args[0];
}
if (args.Length >= 2)
{
baud = int.Parse(args[1]);
}
InitializeComPort(port, baud);
string text;
do
{
String[] mystring = System.IO.Ports.SerialPort.GetPortNames();
text = Console.ReadLine();
int STX = 0x2;
int ETX = 0x3;
ComPort.Write(Char.ConvertFromUtf32(STX) + text + Char.ConvertFromUtf32(ETX));
} while (text.ToLower() != "q");
}
private static void InitializeComPort(string port, int baud)
{
ComPort = new SerialPort(port, baud);
ComPort.PortName = port;
ComPort.BaudRate = baud;
ComPort.Parity = Parity.None;
ComPort.StopBits = StopBits.One;
ComPort.DataBits = 8;
ComPort.ReceivedBytesThreshold = 9;
ComPort.RtsEnable = true;
ComPort.DtrEnable = true;
ComPort.Handshake = System.IO.Ports.Handshake.XOnXOff;
ComPort.DataReceived += OnSerialDataReceived;
OpenPort(ComPort);
}
public static void OpenPort(SerialPort ComPort)
{
try
{
if (!ComPort.IsOpen)
{
ComPort.Open();
}
}
catch (Exception e)
{
throw e;
}
}
}
}
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我的问题是DataReceived事件永远不会被触发.
我的计划规范是:
但我仍然不知道为什么DataReceived事件没有被解雇.
不幸的是,我仍然无法DataReceived以任何方式解雇事件.
所以,我创建了一个新项目,希望我将面临一个解决方法.
在那个新项目[只是控制台应用程序],我创建了一个类......
public class MyTest
{
public SerialPort SPCOM4;
public MyTest()
{
SPCOM4 = new SerialPort();
if(this.SerialPortOpen(SPCOM4, "4"))
{
this.SendToPort(SPCOM4, "com test...");
}
}
private bool SerialPortOpen(System.IO.Ports.SerialPort objCom, string portName)
{
bool blnOpenStatus = false;
try
{
objCom.PortName = "COM" + portName;
objCom.BaudRate = 9600;
objCom.DataBits = 8;
int SerParity = 2;
int SerStop = 0;
switch (SerParity)
{
case 0:
objCom.Parity = System.IO.Ports.Parity.Even;
break;
case 1:
objCom.Parity = System.IO.Ports.Parity.Odd;
break;
case 2:
objCom.Parity = System.IO.Ports.Parity.None;
break;
case 3:
objCom.Parity = System.IO.Ports.Parity.Mark;
break;
}
switch (SerStop)
{
case 0:
objCom.StopBits = System.IO.Ports.StopBits.One;
break;
case 1:
objCom.StopBits = System.IO.Ports.StopBits.Two;
break;
}
objCom.RtsEnable = false;
objCom.DtrEnable = false;
objCom.Handshake = System.IO.Ports.Handshake.XOnXOff;
objCom.Open();
blnOpenStatus = true;
}
catch (Exception ex)
{
throw ex;
}
return blnOpenStatus;
}
private bool SendToPort(System.IO.Ports.SerialPort objCom, string strText)
{
try
{
int STX = 0x2;
int ETX = 0x3;
if (objCom.IsOpen && strText != "")
{
objCom.Write(Char.ConvertFromUtf32(STX) + strText + Char.ConvertFromUtf32(ETX));
}
}
catch (Exception ex)
{
throw ex;
}
return true;
}
}
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我不确定我是好运还是运气不好,因为这个新类可能会产生火灾DataReceived事件,这个事件来自仍在运行的旧控制台应用程序.对我来说这是奇迹,我不知道这是怎么发生的.
让我告诉你更多细节,以便你可以给我更好的建议.
DataReceived在第一个项目中生成fire 事件,同时两个项目正在运行.有人可以给我一些建议,我怎么能将这两个项目合并为一个项目?
Han*_*ant 19
ComPort.Handshake = Handshake.None;
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问题不在于DataReceived事件没有触发,问题是串口没有接收到任何数据.很少有串口设备根本不使用握手.如果将其设置为"无",则驱动程序将不会打开DTR(数据终端就绪)和RTS(请求发送)信号.哪个串口设备解释为"机器已关闭(DTR)"或"机器未准备好接收数据(RTS)".因此它不会发送任何内容,您的DataReceived事件也不会触发.
如果您确实需要None,则将DTREnable和RTSEnable属性设置为true.但是你可能想要HandShake.RequestToSend,因为设备似乎正在关注握手信号.
如果仍然遇到问题,请使用Putty或HyperTerminal等其他串口程序,以确保连接和通信参数良好,并且设备具有响应能力.SysInternals的PortMon实用程序提供了驱动程序交互的低级视图,因此您可以比较好与坏.