Question

我在运行此代码的任何解决方案时都会遇到错误？

{
pinMode(button2pin, INPUT, 3);
  button1State= digitalRead(button1Pin 2);
}
  if (button1State == LOW)
{

}

Answer 1

此代码假定您已将交换机的一侧连接到数字引脚3，另一侧接地，并且您的按钮常开。

const int button2pin = 3;
int button2State = 0;

void setup(){

    pinMode(button2pin, INPUT_PULLUP);
    Serial.begin(9600); // for debug only

}

void loop(){

    button2State = digitalRead(button2pin);

    if(button2State == LOW){

        Serial.print("Button 2 pressed"); // for debug only

    }

}

Answer 2

按下按钮时，此项目会打开灯。也许可以帮助你解决问题。

int buttonPin = 2;
int ledPin = 5;

//This var read the state of Buttonpin

int buttonState = 0;

void setup() {

// set the pin as output
pinMode(ledPin , OUTPUT);

// set the pin as input
pinMode(buttonPin , INPUT);
}

void loop(){
// read the button state

buttonState = digitalRead(buttonPin );

if (buttonState == HIGH) {
    digitalWrite(ledPin, HIGH);
}
else {
    digitalWrite(ledPin, LOW);
}
}

Arduino简单按钮激活灯

2 个答案: