Question

我试图用Windows窗体中的C＃程序连接到arduino。我可以将数据从c＃发送到arduino，但我想从arduino获取数据到C＃程序。

我已经尝试过SerialDataReceivedEventHandler，但是在构建表单后我没有得到数据......

我该怎么办？

       public Form1()
    {
        InitializeComponent();
        Init();

    }//end form 1

    private void Init()
    {
        try
        {
            arduinoPort = new SerialPort();
            arduinoPort.BaudRate = 9600;
            arduinoPort.PortName = "COM4";
            arduinoPort.Handshake = Handshake.None;
            arduinoPort.RtsEnable = true;//request to send true
            arduinoPort.DtrEnable = true;//arduino can send messages to the c# program
            arduinoPort.DataReceived += new SerialDataReceivedEventHandler(GetFromArduino);               
            arduinoPort.Open();
        }//end try

        catch (Exception ex) { MessageBox.Show(ex.Message); }

    }//end init


    private void GetFromArduino(object sender, SerialDataReceivedEventArgs e)
    {
        //string arduinoInputString = arduinoPort.ReadLine();
        //Invoke(new Action(() => label1.Text = arduinoInputString));
        MessageBox.Show("does it work?");
    }//end get from arduino

Answer 1

MessageBox.Show无法从DataReceived活动

开始工作

SerialPort.DataReceived Event

当数据出现时，在辅助线程上引发DataReceived事件从SerialPort对象接收。因为这个事件是在一个辅助线程，而不是主线程，试图修改一些主线程中的元素，如UI元素，可以引发一个线程异常。如果有必要修改main中的元素表单或控件，使用Invoke发回更改请求，这样做关于正确线程的工作

充其量你需要做这样的事情

this.Invoke(new Action(() => { MessageBox.Show(this, "text"); }));

但是，如果您真的想知道该事件是否被触发，那么请使用断点

Using Breakpoints

最后，如果事件未被触发，则您必须查阅设备的文档，以获取设备和串行端口的相应配置。

Answer 2

制作一些处理程序方法，例如

        private void DataReceivedHandler(object sender, SerialDataReceivedEventArgs e)
        {
             SerialPort comm = (SerialPort)sender;
             string incoming_Data = comm.ReadExisting();
             this.BeginInvoke((MethodInvoker)delegate()
             {
                Console.WriteLine(incoming_Data + "\n");
             });          
        }

然后，将此方法发送给 DataRecieved 事件的订阅者

arduinoPort.DataReceived += DataReceivedHandler;

来源：https://social.msdn.microsoft.com/Forums/vstudio/en-US/bd8f7ac8-67d5-4eb5-b679-c595f9c7536d/how-to-print-out-text-from-serialport-datareceived-event?forum=netfxbcl

Answer 3

I don't know if it matters, but this what i did and it solved:
1. in the setup function of the arduino program i had a loop that ran on all the pins, now i specified it to the pins i really use.
2. i don't open the port in the form.cs, but in the program.cs and run a while loop that just check if the port is open. then, it gets the data without disturbing the form to run... 

**in the Program.cs:**


  public static SerialPort arduinoPort { get; set; }
    public static string arduinoInputString { get; set; }
    [STAThread]

    static void Main()
    {
        arduinoPort = new SerialPort();
        try
        {
            arduinoPort = new SerialPort();
            arduinoPort.BaudRate = 250000;
            arduinoPort.PortName = "COM4";
            //arduinoPort.Handshake = Handshake.None;
            //arduinoPort.RtsEnable = true;//request to send true
            //arduinoPort.DtrEnable = true;//arduino can send messages to the c# program
            ////arduinoPort.DataReceived += new SerialDataReceivedEventHandler(GetFromArduino);
            arduinoPort.DataReceived += DataReceivedHandler;
            arduinoPort.Open();
        }//end try
        catch (Exception ex) { Console.WriteLine((ex.Message)); }

        Application.EnableVisualStyles();
        Application.SetCompatibleTextRenderingDefault(false);
        Application.Run(new Form1());

        while (arduinoPort.IsOpen)//read data if the port is open
        {

        }



    }//end main

    static void DataReceivedHandler(object sender, SerialDataReceivedEventArgs e)
    {
        SerialPort comm = (SerialPort)sender;
        arduinoInputString = comm.ReadExisting();
        Form1.label1.Text = arduinoInputString;//label to show the input string from arduino
    }//end get data from arduino

SerialDataReceivedEventHandler不工作c＃

3 个答案: