串口接收数据

Question

       初学C#，为什么第一个时间错误。  第二个事件就可以啊？
        private void serialPort_DataReceived(object sender, System.IO.Ports.SerialDataReceivedEventArgs e)
        {
            byte[] rsvdata = Convert.FromBase64String(serialPort.ReadLine());//读取输入缓冲区
            string str = Encoding.Unicode.GetString(rsvdata);                //解码
            serialPort.Close();//关闭串口            rsvbox.Text = str;//错误！
        }        private void button1_Click(object sender, EventArgs e)
        {
            byte[] rsvdata = Convert.FromBase64String(serialPort.ReadLine());//读取输入缓冲区
            rsvbox.Text =  Encoding.Unicode.GetString(rsvdata);              //解码
            serialPort.Close();                                              //关闭串口*/
        }

Answer 1

        void serialPort_DataReceived(object sender, SerialDataReceivedEventArgs e)
        {
            int count = upsCom.BytesToRead;            byte[] buffer = new byte[count];            serialPort.Read(buffer, 0, count);            string resultString = Encoding.Default.GetString(buffer);//这里要根据实际情况选择解码            if (serialPort.IsOpen)
            {
                serialPort.DiscardInBuffer();
            }
        }

Answer 2

更正下：
        void serialPort_DataReceived(object sender, SerialDataReceivedEventArgs e)
        {
            int count = serialPort.BytesToRead;            byte[] buffer = new byte[count];            serialPort.Read(buffer, 0, count);            string resultString = Encoding.Default.GetString(buffer);//这里要根据实际情况选择解码            if (serialPort.IsOpen)
            {
                serialPort.DiscardInBuffer();
            }
        }

Answer 3

rsvbox.Text = str;//错误！
这里跨线程调用了。提示应该是不是在创建rsvbox的线程上访问它。如果是的话请改用安全的异步调用方法，比如Invoke或用BackgroundWorker异步编程。