Question

功能性：

在按下红色圆顶按钮（不是2状态按钮）之前，串行监视器将打印一个列表＆＃34; 0＆＃34; s，当按下红色圆顶按钮时，按钮状态将从LOW切换为高，因此在串行监视器上将打印一个＆＃34; 1＆＃34; s列表。

但是，当按钮处于HIGH状态时，串行监视器打印＆＃34; 1＆＃34; s和用户将无法将按钮状态从HIGH切换到LOW。因此，按钮只能在一段时间（25秒）后自动从HIGH切换到LOW。

因此， 正确行为 ：

初始状态print =＆gt; 00000000000000（当用户按下红色圆顶按钮=＆gt; buttonstate将LOW更改为HIGH时）111111111111111111（当用户按下按钮时，没有任何反应）111111111111111111（25秒延迟后，buttonState将切换为HIGH到LOW）0000000000000

问题：

此时，用户可以在LOW到HIGH和HIGH到LOW之间切换。意思是，flow =＆gt; 00..000（用户按下按钮，切换到高电平）111 ... 111（用户按下按钮，切换到高电平为低电平）0000 ...

我不确定如何启用按钮仅从LOW切换到HIGH但禁用按钮从HIGH切换到LOW。

当用户按下按钮时，它可以改变按钮状态，从＆＃34; 0＆＃34;到＆＃34; 1＆＃34;但是当它处于＆＃34; 1＆＃34;。

时无法改变按钮状态

因此，我想请求一些允许以下正确行为的帮助。

由于

代码：

const int buttonPin = 2; //the number of the pushbutton pin
const int Relay     = 4; //the number of the LED relay pin

uint8_t    stateLED = LOW;
uint8_t      btnCnt = 1;

int buttonState = 0; //variable for reading the pushbutton status
int buttonLastState = 0;
 int outputState = 0;

void setup() {
  Serial.begin(9600);
  pinMode(buttonPin, INPUT); 
  pinMode(Relay, OUTPUT);
  digitalWrite(Relay, LOW);
}

void loop() {

  // read the state of the pushbutton value:
   buttonState = digitalRead(buttonPin);
  // Check if there is a change from LOW to HIGH
  if (buttonLastState == LOW && buttonState == HIGH)
  {
     outputState = !outputState; // Change outputState
  }
  buttonLastState = buttonState; //Set the button's last state

  // Print the output
  if (outputState)
  {
     switch (btnCnt++) {
      case 100:
         stateLED = LOW;
        digitalWrite(Relay, HIGH); // after 10s turn on
        break;

      case 250:
        digitalWrite(Relay, LOW); // after 20s turn off
        //Toggle ButtonState to LOW from HIGH without user pressing the button
        digitalWrite(buttonPin, LOW);
        break;

      case 252: // small loop at the end, to do not repeat the LED cycle
        btnCnt--;
         break;    
    }

    Serial.println("1");
  }else{
    Serial.println("0");
    if (btnCnt > 0) {  
      // disable all:
      stateLED = LOW;
      digitalWrite(Relay, LOW);
    }
    btnCnt = 0;
  }

  delay(100);
}

Answer 1

您需要设置outputState并让它设置，直到它在25秒后重置。如果仍按下按钮，则它将循环显示251。

const int buttonPin = 2; //the number of the pushbutton pin
const int Relay     = 4; //the number of the LED relay pin

uint8_t    stateLED = LOW;
uint8_t      btnCnt = 1;

bool outputState = false;

void setup() {
  Serial.begin(9600);
  pinMode(buttonPin, INPUT); 
  pinMode(Relay, OUTPUT);
  digitalWrite(Relay, LOW);
}

void loop() {

  outputState |= digitalRead(buttonPin); // if pushButton is high, set outputState (low does nothing)

  // Print the output
  if (outputState)
  {
     switch (btnCnt++) {
      case 100:
         stateLED = LOW;
        digitalWrite(Relay, HIGH); // after 10s turn on
        break;

      case 250:
        digitalWrite(Relay, LOW); // after 20s turn off
        //Toggle ButtonState to LOW from HIGH without user pressing the button
        outputState = false; // reset state to low
        break;
      case 251: // loop (it might happen, if previous step sets outputState=false but button is still pressed -> no action)
        --btnCnt;
        outputState = false;
        break;
    }

    Serial.println("1");
  }else{
    Serial.println("0");
    if (btnCnt > 0) {  
      // disable all:
      stateLED = LOW;
      digitalWrite(Relay, LOW);
    }
    btnCnt = 0;
  }

  delay(100);
}

防止ButtonState从HIGH切换到LOW

1 个答案: