Question

我想打印＆＃34; Entry＆＃34;在串行监视器中每次传感器感应到物体，但我的输出是＆＃34;进入＆＃34; ＆＃34;条目＆＃34;等等，直到感觉到下一个条目并且它继续以这种格式。我想打印＆＃34;条目＆＃34;只有当感测到物体的条目时才会这样。

`

#define trig1 11
#define echo1 10
#define trig2 9
#define echo2 6
#define ledpin 13
int sensor_1 = 0;
int sensor_2 = 0;
int count = 0;
long duration1, distance1;
long duration2, distance2;

//unsigned long prevmillis=0;
//const long interval=1000;


void setup() {
  // put your setup code here, to run once:
  pinMode(trig1, OUTPUT);
  pinMode(echo1, INPUT);
  pinMode(trig2, OUTPUT);
  pinMode(echo2, INPUT);
  pinMode(ledpin, OUTPUT);
  Serial.begin(9600);

}

void loop() {
  //unsigned long prevmillis=millis();

  sensor1(); ///set sensor_1=1
  sensor2();///set sensor_2=1
  if (sensor_1 == 1)
  { if (sensor_2 == 1) {
      Serial.println("Entry");

    }
  }


}

int sensor1(void)
{
  digitalWrite(trig1, LOW);
  delayMicroseconds(2);
  digitalWrite(trig1, HIGH);
  delayMicroseconds(10);
  digitalWrite(trig1, LOW);
  duration1 = pulseIn(echo1, HIGH);

  analogWrite(ledpin, 0);
  distance1 = duration1 * .034 / 2;
  if (distance1 < 10) {
    delay(100);
    sensor_1 = 1;

    //delay(400);

    //count++;
    //unsigned long currentmillis=millis();

    //Serial.println(currentmillis-prevmillis);
    //Serial.println("Sensor1 Sensed Object");
    //Serial.println(count);
    analogWrite(ledpin, 255);
    delay(200);

  }

  //return sensor_1;
}
int sensor2(void)
{
  digitalWrite(trig2, LOW);
  delayMicroseconds(2);
  digitalWrite(trig2, HIGH);
  delayMicroseconds(10);
  digitalWrite(trig2, LOW);
  duration2 = pulseIn(echo2, HIGH);

  analogWrite(ledpin, 0);
  distance2 = duration2 * .034 / 2;
  if (distance2 < 15) {
    delay(100);
    sensor_2 = 1;

    //delay(400);

    //count++;
    //unsigned long currentmillis=millis();

    //Serial.println(currentmillis-prevmillis);
    //Serial.println("Sensor2 Sensed Object");
    //Serial.println(count);
    analogWrite(ledpin, 255);
    delay(100);

  }
  //return sensor_2;
}

`

Answer 1

使用后您无法清除传感器条件值。

if (sensor_1 == 1)
{ 
  if (sensor_2 == 1) 
  {
    Serial.println("Entry");
    sensor_1 = 0;
    sensor_2 = 0;
  }
}

应该做的伎俩。

如何连续打印＆＃34; Entry＆＃34;每次传感器感知对象条目

1 个答案: