Question

我编写了一个处理MQTT消息（有效负载）的回调函数。我想切换引脚而不是基于传入的有效载荷。我现在遇到的问题是第一个语句在发送11时执行得很好。但是第二个语句不起作用。奇怪的是，当我发送两条包含1x和另一条x1的消息时，我可以再次切换两个引脚。这是一个非常奇怪的问题！非常感谢任何人都可以帮助我。我正在使用PubSubClient库。

 void callback(char* topic, byte* payload, unsigned int length) {
      Serial.print("Message arrived [");
      Serial.print(topic);
      Serial.print("] ");
      for (int i = 0; i < length; i++) {
        Serial.print((char)payload[i]);
      }

       //Set GPIO0 to HIGH or LOW on first character received in message
      if (payload[0] == '1') {
        digitalWrite(GPIO0, HIGH);   // Turn the relay on
        client.publish(getOutTopic().c_str(), "GPIO0 set to HIGH");
      } else if (payload[0] == '0') {
        digitalWrite(GPIO0, LOW);  // Turn the relay off
        client.publish(getOutTopic().c_str(), "GPIO0 set to LOW");
      }

      //Set GPIO2 to HIGH or LOW on first character received in message
      if (payload[1] == '1') {
        digitalWrite(GPIO2, HIGH);   // turn LED off. With High it is inactive  on the ESP-01)
        client.publish(getOutTopic().c_str(), "GPIO2 set to HIGH");
      } else if (payload[1] == '0') {
        digitalWrite(GPIO2, LOW);  // Turn the LED on by making the voltage LOW
        client.publish(getOutTopic().c_str(), "GPIO2 set to LOW");
      } 
}

Answer 1

client.publish破坏有效负载缓冲区。所以我必须将有效载荷捕获到新的变量中才能解决问题。

 void callback(char* topic, byte* payload, unsigned int length) {
      char p0 = (char)payload[0];
      char p1 = (char)payload[1];
      Serial.print("Message arrived [");
      Serial.print(topic);
      Serial.print("] ");
      for (int i = 0; i < length; i++) {
        Serial.print((char)payload[i]);
      }

       //Set GPIO0 to HIGH or LOW on first character received in message
      if (p0 == '1') {
        digitalWrite(GPIO0, HIGH);   // Turn the relay on
        client.publish(getOutTopic().c_str(), "GPIO0 set to HIGH");
      } else if (p0 == '0') {
        digitalWrite(GPIO0, LOW);  // Turn the relay off
        client.publish(getOutTopic().c_str(), "GPIO0 set to LOW");
      }

      //Set GPIO2 to HIGH or LOW on first character received in message
      if (p1 == '1') {
        digitalWrite(GPIO2, HIGH);   // turn LED off. With High it is inactive  on the ESP-01)
        client.publish(getOutTopic().c_str(), "GPIO2 set to HIGH");
      } else if (p1 == '0') {
        digitalWrite(GPIO2, LOW);  // Turn the LED on by making the voltage LOW
        client.publish(getOutTopic().c_str(), "GPIO2 set to LOW");
      } 
}

回调函数不执行第二个语句 - C ++ MQTT

1 个答案: