Question

我有以下代码需要从端口读取数据然后在文本框中显示。我为此目的使用DataReceived事件处理程序，但不知道如何在文本框中显示此数据。从各种来源我了解到Invoke方法应该用于此，但不知道如何使用它。建议请...

    private void Form1_Load(object sender, EventArgs e)
    {
        //SerialPort mySerialPort = new SerialPort("COM3");
        mySerialPort.PortName = "COM3";
        mySerialPort.BaudRate = 9600;
        mySerialPort.Parity = Parity.None;
        mySerialPort.StopBits = StopBits.One;
        mySerialPort.DataBits = 8;
        mySerialPort.Handshake = Handshake.None;
        mySerialPort.DataReceived += new SerialDataReceivedEventHandler(mySerialPort_DataReceived);
        mySerialPort.Open();
    }

    private void mySerialPort_DataReceived(object sender, SerialDataReceivedEventArgs e)
    {
        SerialPort sp = (SerialPort)sender;
        string s= sp.ReadExisting();
        // next i want to display the data in s in a textbox. textbox1.text=s gives a cross thread exception
    }
    private void button1_Click(object sender, EventArgs e)
    {

        mySerialPort.WriteLine("AT+CMGL=\"ALL\"");

    }

Answer 1

MSDN包含一个好的article，其中包含有关使用其他线程的控制方法和属性的示例。

简而言之，您需要的是一个委托方法，它使用给定的字符串设置文本框的Text属性。然后，通过mySerialPort_DataReceived方法从TextBox.Invoke()处理程序中调用该委托。像这样：

public delegate void AddDataDelegate(String myString);
public AddDataDelegate myDelegate;

private void Form1_Load(object sender, EventArgs e)
{
    //...
    this.myDelegate = new AddDataDelegate(AddDataMethod);
}

public void AddDataMethod(String myString)
{
    textbox1.AppendText(myString);
}

private void mySerialPort_DataReceived(object sender, SerialDataReceivedEventArgs e)
{
   SerialPort sp = (SerialPort)sender;
   string s= sp.ReadExisting();

   textbox1.Invoke(this.myDelegate, new Object[] {s});       
}

Answer 2

试试这个（适合我）：

private delegate void UpdateUiTextDelegate(string text);

private void Receive(object sender, System.IO.Ports.SerialDataReceivedEventArgs e)
{
    if (mySerialPort.IsOpen)
    {
        RxString = mySerialPort.ReadLine();
        Dispatcher.Invoke(DispatcherPriority.Send, new UpdateUiTextDelegate(DisplayText), RxString);
    }
}

private void DisplayText(string RxString)
{
    myTextBox.Text = RxString;
}

Answer 3

我正在创建一个GUI＆＃34; Form＆＃34;用于USB COM端口。这就是我如何将数据发送到窗口而不会得到一个＆＃34;交叉线程＆＃34;错误。

private void serialPort1_DataReceived(object sender, SerialDataReceivedEventArgs e)
{
    string inData = serialPort1.ReadLine(); // ReadLine includes the + "\n"
    displayToWindow(inData);
}

private void displayToWindow(string inData)
{
    BeginInvoke(new EventHandler(delegate
    {
        richTextBox1.AppendText(inData);
        richTextBox1.ScrollToCaret();
    }));
}

如何显示在serialport的DataReceived事件处理程序中读取的数据

3 个答案: