C#代码
using System;
using System.Windows.Forms;
using System.IO.Ports;
SerialPort port;
private void btnStart_Click(object sender, EventArgs e)
{
port = new SerialPort("COM6", 9600);
port.Open();
port.Write("START");
port.Close();
}
Arduino代码
"#"include "MOVIShield.h"
MOVI recognizer(true);
循环内的代码
signed int res=recognizer.poll();
if(Serial.available() > 0){
String data = Serial.readString();
if(data = "START"){
recognizer.ask("Hello. My name is John");
}
}
我尝试点击btnStart发送" START"到我的Arduino程序和Arduino程序应该在收到C#程序的数据后运行询问("你好。我的名字是John")。但是当我点击btnStart时,什么也没发生。
答案 0 :(得分:1)
你可以尝试不同的几件事:
1-确保两端的COM端口参数相同
如http://www.instructables.com/id/How-to-connect-Arduino-to-a-PC-through-the-serial-/
所述将其添加到循环外的Arduino C代码中:
Serial.begin(9600);
将您的C#更改为类似于以下的代码:
private void btnStart_Click(object sender, EventArgs e)
{
port = new SerialPort("COM6", 9600);
port.DataBits = 8;
port.StopBits = StopBits.One;
port.Handshake = Handshake.None;
port.Parity = Parity.None;
port.Open();
port.Write("START");
port.Close();
}
2-使用与C#不同的工具来测试您是否可以与Arduino进行通信。
e.g。这个工具有15天的演示:https://www.eltima.com/rs232-testing-software/
答案 1 :(得分:0)
我想这行中的单一等于可能与它有关。 if(data =“START”)
答案 2 :(得分:0)
也许您应该使用PortWrite();
而不是port.write();
这是一个可以帮助您的类似演示:
C#代码:
using System;
using System.Windows.Forms;
using System.IO.Ports;
namespace lightcontrol
{
public partial class Form1 : Form
{
SerialPort port;
public Form1()
{
InitializeComponent();
this.FormClosed += new FormClosedEventHandler(Form1_FormClosed);
if (port==null)
{
port = new SerialPort("COM7", 9600);//Set your board COM
port.Open();
}
}
void Form1_FormClosed(object sender,FormClosedEventArgs e)
{
if(port !=null &&port.IsOpen)
{
port.Close();
}
}
private void button1_Click(object sender, EventArgs e)
{
PortWrite("1");
}
private void button2_Click(object sender, EventArgs e)
{
PortWrite("0");
}
private void PortWrite(string message)
{
port.Write(message);
}
}
}
Arduino代码:
const int LedPin = 13;
int ledState = 0;
void setup()
{
pinMode(LedPin, OUTPUT);
Serial.begin(9600);
}
void loop()
{
char receiveVal;
if(Serial.available() > 0)
{
receiveVal = Serial.read();
if(receiveVal == '1')
ledState = 1;
else
ledState = 0;
}
digitalWrite(LedPin, ledState);
delay(50);
}
这是教程: http://www.lattepanda.com/topic-f6t1534p6387.html?sid=8abb18a037a0308db9e4d5825a0aebcd#p6387