按钮发送命令 SerialPort1_DataReceived (EventArgs e)

Question

//Button need DataRecieved  event handler

private void LoggedDataBtn_Click(object sender, EventArgs e)

{

    //When LoggDataBtn is pressed it Send the command LogToShow to  device

    string LoggedDataTOshow = "?R\r\n";

    string commForMeter = string.Format(LoggedDataTOshow);

    try

    {

        if (serialPortForApp.IsOpen)

        {

            Thread.Sleep(2000);

            serialPortForApp.Write(LoggedDataTOshow);

        }

    }

    catch (Exception)

    {

        ShowDataInScreenTxtb.Text = "TimeOUT Exception: Error for requested command";

    }

}

//button does not need DataRecieved handler to display but create conflict 

//for those that need it

private void GLPBtn_Click(object sender, EventArgs e)

{

    //send command to instrument

    string GLPCommand = "?G\r\n";

    string GLPToShow = String.Format(GLPCommand);

    try

    {

        if (serialPortForApp.IsOpen)

        {

            Thread.Sleep(2000);

            //write command to port

            serialPortForApp.Write(GLPToShow);

        }

    }

    catch

    {

        //MessageBox error if data reading is not posible

        ShowDataInScreenTxtb.Text = "TimeOUT Exception";

    }

    while (glpShowInMainForm.DataToSetandGet != "ENDS\r")

    {

        serialPortForApp.Write("\r");

        DataToSetandGet = serialPortForApp.ReadExisting();

        ShowDataInScreenTxtb.AppendText(DataToSetandGet.Replace("\r", "\r\n"));

    }

}

//如何创建开关盒或定义不同的按钮值以读取和显示。一些按钮与 datarecived 事件一起工作，而其他按钮不需要它，但会为那些需要在 datarecieved 事件处理程序（如我的 LoggedDataBtn_Click ）中读取退出的按钮产生冲突，其他按钮需要将 ReadExisted 放入 button_click 方法中，而不是像我的 GLPBtn_Click 那样使用 datarecieved 事件.