Question

我正在尝试编写一个将执行以下操作的程序：

编写代码，如果1然后输入按下或发送点击，然后再次输入1并输入按下或发送点击导致LED 1打开，如果以类似的方式输入'1''0'则LED 1转动关闭，等等LED 2和3，即：'2''1'打开LED 2，'3''0'关闭LED 3。

到目前为止，这是我的代码：

int incomingVal;
int ledPin = 16; 
int ledPin2 = 15; 
int ledPin3 = 14; 

void setup()
{

  Serial.begin(9600);     // opens serial port, sets data rate to 9600 bps
  Serial.println("starting");
  pinMode(ledPin,OUTPUT);
  pinMode(ledPin2,OUTPUT);
  pinMode(ledPin3,OUTPUT);
}


void checkForRecvdChar (); 

void loop()
{
  if (Serial.available() > 0 ) //then chars are in the serial buffer
  {
    incomingVal = Serial.parseInt();
    Serial.print("You entered: ");
    Serial.println(incomingVal);

    if (incomingVal == 10);//turns off led 1 
    {
      digitalWrite(ledPin, LOW);  

    }

     if (incomingVal == 11);//turns on led 1 
    {
      digitalWrite(ledPin, HIGH);  

    }

     if (incomingVal == 20);//turns off led 2
    {
      digitalWrite(ledPin2, LOW);  

    }

     if (incomingVal == 21);//turns on led 2
    {
      digitalWrite(ledPin2, HIGH);  

    }

     if (incomingVal == 30);//turns off led 3
    {
      digitalWrite(ledPin3, LOW);  

    }

     if (incomingVal == 31);//turns on led 3
    {
      digitalWrite(ledPin3, HIGH);  

    }

  }

}

现在无论我输入什么组合，它都会打开所有LED

Answer 1

你的if语句后面有半冒号，终止了if语句。因此，所有预期的if块都会被执行。

例如......

if (incomingVal == 31);//turns on led 3
    {
      digitalWrite(ledPin3, HIGH);  

    }

需要

  if (incomingVal == 31)
    {
      digitalWrite(ledPin3, HIGH);  

    }

Answer 2

 Ok So First It Will Better To Use Switch Case in this Case:
1. You Can Just Use The Same Number To Turn The Led ON and OFF 
Example:

    case 1:
        {
            if(wFlag == false)
            {
               digitalWrite(LED, HIGH);
               wFlag = true;
               Serial.println("LED is ON");
            }
            else if (wFlag == true)
            {
                digitalWrite(LED, LOW);
                wFlag = false;
                Serial.println("LED is OFF");
            }
            break;
         }

2. The Way You Do It, Your Problem Is Serial Read Gets Only Byte after Byte
   So When You Press 1 or 11 It Sees 1 Because It is The First one "1" and Then Do Your If             
   Statment.
   You Can Fix it Like This: 

        while (Serial.available())
        {
            IncomingData = Serial.parseInt();
             Temp += String(IncomingData);    //Temp String Var
        }
       DataIN = Temp.toInt();  //DataIN int Var
       Temp ="";                         // Rest The Temp String
        Switch(DataIN)           //  Or if(DataIN == 1)
        {

        }

使用串口打开/关闭3个LED

2 个答案: