C＃和arduino连接 - 控制WPF应用程序 [英] C# and arduino connection - controling a WPF application

问题描述

我仍然是编程世界的新手，但如果可能，我需要一些帮助。如果没有，那也没关系。

我有这个任务，我必须在C＃（WPF）和Arduino之间建立连接。现在，我设法连接了液晶显示器，因此它将电位计的值输出到液晶显示屏，如图片一旦我在Arduino和C＃之间建立连接，C＃必须将这些值从电位计输出到Line表单。所以，用简短的话说，我如何用电位计改变当前值，C＃中的线必须像我一样移动。我非常擅长解释所以希望你明白我在说什么：P如果有人能帮助我完成这项任务，我会非常高兴。提前致谢



节目：Microsoft Visual Studio（2008-2015），Arduino（最新版）

语言：C＃

应用程序：WPF（.NET Framework 4.5）

操作系统：Windows



行表单代码：

I'm still new to the world of programming but I need some help with this if it's possible. If not, that's okay too.
I have this task where I have to establish a connection between C# (WPF) and Arduino. Now, I managed to connect the LCD so it outputs the values from the potenciometer to the LCD screen like in this Picture Once I have the connection between Arduino and C#, C# must output those values from the potenciometer to the "Line" form. So, in shorter words, How I change the current value with the potenciometer, the line in C# MUST be moving as I do it. I'm very bad at explaining so hopefully You understood what I was saying :P If anyone could help me with this task, I'd be more than happy. Thanks in advance

Programs: Microsoft Visual Studio (2008-2015), Arduino (latest version)
Language: C#
Application: WPF (.NET Framework 4.5)
OS: Windows

Line Form code:

    <Line
    Name="TheLine"
    Stretch="Fill"
      X1="1" X2="1" Y1="0" Y2="1"
      Width="2"
      HorizontalAlignment="Center"
      Stroke="Black"
      >
    <Line.RenderTransform>
        <TransformGroup>
            <RotateTransform CenterX="1" CenterY="{Binding ActualHeight, ElementName=TheLine}"

                              Angle="{Binding Path=Value, ElementName=AngleSlider}" />
        </TransformGroup>
    </Line.RenderTransform>
</Line>
<StackPanel Grid.Column="1">
    <Slider Name="AngleSlider" Minimum="-90" Maximum="90" Value="0" Height="59" Width="200" />
</StackPanel>

Arduino代码：

Arduino code:

#include <LiquidCrystal.h>

 // initialize the library with the numbers of the interface pins
LiquidCrystal lcd(12, 11, 5, 4, 3, 2);

int sensorPin = A0;    // select the input pin for the potentiometer
int sensorValue = 0; // variable to store the value coming from the sensor

 
 void setup() {
  // set up the LCD's number of columns and rows: 
  lcd.begin(16, 2);
  // Print a message to the LCD.
  lcd.print("Welcome!");
  delay(3000);
  lcd.clear();
  Serial.begin(9600);  
}

 void loop() {
  lcd.print("Current value:");
  // read the value from the sensor:
  sensorValue = analogRead(sensorPin);     
  lcd.setCursor(0, 1);
  lcd.print(sensorValue /4);
  lcd.print("/255 ");
  delay(10);
  lcd.clear();
  Serial.println(sensorValue);
  delay(10);
  
}

我的尝试：



我设法在c＃和arduino之间建立连接但是从电位计读取值并将它们带到线上就是我的奋斗。

What I have tried:

I managed to establish a connection between c# and arduino but to read the values from the potenciometer and bring them onto the line is what I struggle.

推荐答案

你很顺利。您接下来需要的只是本文。当然你使用的是Arduino而不是PIC。使用该示例作为解决方案的模板。



使用WPF，RS232和PIC通信的串行通信 [ ^ ]

You are well on the way. All you need next is in this article. Of course you are using Arduino and not PIC. Use the example as a template for your solution.

Serial Communication using WPF, RS232 and PIC Communication[^]

请参阅我对该问题的评论。 />


对于PC端，带.NET的Windows，请参阅开源项目libusb和LibUsbDotNet：

libusb ，

LibUsbDotNet 。



在Arduino方面，您可以从这里开始： Arduino - SoftwareSerial 。

-SA

Please see my comments to the question.

For PC side, Windows with .NET, please see the open-source projects libusb and LibUsbDotNet:
libusb,
LibUsbDotNet.

On Arduino side, you can start here: Arduino — SoftwareSerial.

—SA

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