COM Port data write/read

[jeb2000]

I have a device connected to a COM port. I need to send it a couple of bytes of data and then it returns me a couple of bytes of data - basically I need to poll it for data readings. I just need a basic console application to do this job. I am new to this area and not sure where to start - any pointers would be much appreciated.

Thanks.

[Alex Papadimoulis]

The System.IO.Ports will be able to help you here. The main class in this namespace is SerialPort.

Opening, reading/writing, and closing is pretty easy.

bool continue = true;
SerialPort mySerialPort = new SerialPort();
 
mySerialPort.PortName = "COM1";
mySerialPort.BaudRate = 9800;
 
// ...
 
mySerialPort.Open();
while (continue) {
  string msg = mySerialPort.ReadLine();  
  ProcessMessage(msg);
  Thread.Sleep(100);
}
 
mySerialPort.Close();

[jeb2000]

Alex,

Thank you for the reply. The data I need to send and receive is hex data so I have tried to adapt your suggestion as follows.

mySerialPort.Open();

// read the port
while (readPort)
{
mySerialPort.Write(new byte[] { 0x01, 0x05 }, 0, 2);
// read the data
List<byte> myData= new List<byte>();
while (mySerialPort.BytesToRead > 0)
myData.Add((byte)mySerialPort.ReadByte());

ProcessData(myData);

// clear the data
myData.Clear();

Thread.Sleep(100);
}

mySerialPort.Close();


I know I have to send a 2 byte request and then receive two bytes of data back. However, my code isn't reading any sensible data back. Any thoughts?

Many thanks.

[Alex Papadimoulis]

Difficult to say why "sensible" data isn't coming back without knowing anything about the devices, etc. However, if it's simply a 2-byte request followed by a 2-byte response, I would write code like this:

mySerialPort.Write(new byte[] { 0x01, 0x05 }, 0, 2);
 
while (mySerialPort.BytesToRead < 2) Thread.Sleep(100);
 
byte[] response = new byte[2];
mySerialPort.Read(response, 0, 2);

[jeb2000]

Alex,

Your suggestion works perfectly - I am now getting excactly what I expect from my device.

Many thanks for your input.