请教门碰传感器延迟输出告警怎么写

jeesir

想做一个门碰传感器，门开输出告警，当门开状态大于1分钟才产生告警。（门开状态，如果1分钟内门关，不产生告警，并重新计时判断是否超过1分钟）
下面肯定写错了 有人熟悉的吗
void loop() {
  // put your main code here, to run repeatedly:
if (digitalRead(IN3) == HIGH)//关门
  {
   delay(60000);
   digitalWrite(OUT2, LOW);
    }
  else if (digitalRead(IN3) == LOW)//开门
  {
   delay(60000);
   digitalWrite(OUT2, HIGH);
  }
}

XlinliY.Zhang

用millis()计时不就行

奈何col

用millis()+1

i7456

1. 
   
2. int IN3 = 3;
   
3. int OUT2 = 2;            
   
4. 
   
5. unsigned long count = 0;   
   
6. unsigned long previousMillis = 0;        
   
7. const long interval = 1000;           
   
8. 
   
9. void setup()
   
10. {
    
11.   // set the digital pin as output:
    
12.   pinMode(OUT2, OUTPUT);
    
13. }
    
14. 
    
15. void loop()
    
16. {
    
17.   unsigned long currentMillis = millis();
    
18. 
    
19.   if (currentMillis - previousMillis >= interval)
    
20.   {
    
21.    previousMillis = currentMillis;
    
22.    if (digitalRead(IN3) == LOW)//开门
    
23.       count++;
    
24.    else count = 0;
    
25.    }
    
26.    if(count > 60)
    
27.     digitalWrite(OUT2, HIGH);
    
28.    else digitalWrite(OUT2, LOW);
    
29. }

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