本文介绍了我们如何阅读我们从arduino中提取的代码的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我们想要做的是当arduino将信号发送到我们正在制作的C#应用​​程序时,当arduino发送A信号时,会出现一个通知弹出窗口,如果它没有发送,通知框将不会出现。



我尝试了什么:



 {
public partial class Form1:表格
{
public Form1()
{
InitializeComponent();
serialPort1.Open();
}

private void serialPort1_DataReceived( object sender,System.IO.Ports.SerialDataReceivedEventArgs e)
{

if ( serialPort1.Read( A));
{Notification.Enabled = true ; }
else
{Notification.Enabled = false ; }
}

私有 void Notification_Click( object sender,EventArgs e)
{
PopupNotifier popup = new PopupNotifier();
popup.Image = Properties.Resources.info;
popup.TitleText = Alert;
popup.ContentText = 嘿嘿;
popup.Popup();
}
}
}
解决方案

What we want to do is when the arduino will send a signal to the C# app that we are making and when the arduino sent the A signal, a notification popup should appear, if it did not send, the notification box will not appear.

What I have tried:

{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
            serialPort1.Open();
        }

        private void serialPort1_DataReceived(object sender, System.IO.Ports.SerialDataReceivedEventArgs e)
        {

            if (serialPort1.Read("A")) ;
            { Notification.Enabled = true; }
            else
            { Notification.Enabled = false; }
        }

        private void Notification_Click(object sender, EventArgs e)
        {
            PopupNotifier popup = new PopupNotifier();
            popup.Image = Properties.Resources.info;
            popup.TitleText = "Alert";
            popup.ContentText = "Hey";
            popup.Popup();
        }
    }
}
解决方案


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08-20 10:16
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