Arduino简易按钮激活灯
当我运行这段代码时,我总是遇到一个错误。有什么解决方案吗Arduino简易按钮激活灯,arduino,Arduino,当我运行这段代码时,我总是遇到一个错误。有什么解决方案吗 { pinMode(button2pin, INPUT, 3); button1State= digitalRead(button1Pin 2); } if (button1State == LOW) { } 此代码假定您已将开关的一侧连接到数字针脚3,另一侧连接到接地,并且您的按钮处于常开状态 const int button2pin = 3; int button2State = 0; void setup(){
{
pinMode(button2pin, INPUT, 3);
button1State= digitalRead(button1Pin 2);
}
if (button1State == LOW)
{
}
此代码假定您已将开关的一侧连接到数字针脚3,另一侧连接到接地,并且您的按钮处于常开状态
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
}
}
按下按钮时,此项目将打开灯。也许能帮你解决问题
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);
}
}
digitalRead(按钮1针脚2)
语法不正确。在问题中包含实际的错误消息。