【发布时间】:2019-07-26 19:16:43
【问题描述】:
这里是场景。我有一个 esp32、2 个 LED 和一个 ios 示例应用程序,我在网上找到了 here。目前,如果我按下我的 esp32 上的按钮,它会通知 ios 应用程序显示“1”,如果再次按下它会显示“0”。这完美地工作。
棘手的部分是这个 ios 应用程序允许我向 esp32 发送写入命令。我想这样做,如果发送“1”,LED A 亮起,LED B 熄灭,那么当发送 0 时,LED A 熄灭,LED B 亮起。我无法做到这一点。尽我所能,我无法弄清楚这个项目的链条中哪里出了问题。可能是 esp32 上的代码,也可能是我不确定的应用程序。
这是我的 arduino 代码。 (代码还有很多没有提到,我其实有 4 个 LED,但我只想在发送写命令时打开 2 个特定的)。
#include <BLEDevice.h>
#include <BLEServer.h>
#include <BLEUtils.h>
#include <BLE2902.h>
BLEServer* pServer = NULL;
BLECharacteristic* pCharacteristic = NULL;
bool deviceConnected = false;
bool oldDeviceConnected = false;
boolean oldState = LOW;
uint32_t value = 0;
#define SERVICE_UUID "4fafc201-1fb5-459e-8fcc-c5c9c331914b"
#define CHARACTERISTIC_UUID "beb5483e-36e1-4688-b7f5-ea07361b26a8"
class MyServerCallbacks: public BLEServerCallbacks {
void onConnect(BLEServer* pServer) {
deviceConnected = true;
};
void onDisconnect(BLEServer* pServer) {
deviceConnected = false;
}
};
class MyCallbacks: public BLECharacteristicCallbacks {
void onWrite(BLECharacteristic *pCharacteristic) {
std::string rxValue = pCharacteristic->getValue();
if (rxValue.length() > 0) {
Serial.print("Received Value: ");
for (int i = 0; i < rxValue.length(); i++) {
Serial.print(rxValue[i]);
}
Serial.println();
if (rxValue.find("1") != -1) {
digitalWrite(13, HIGH);
digitalWrite(27, LOW);
}
else if (rxValue.find("0") != -1) {
digitalWrite(13, LOW);
digitalWrite(27, HIGH);
}
}
}
};
const int bt1 = 14;
boolean bt1g = true;
int bt1t = 0;
void setup() {
pinMode(13, OUTPUT);
pinMode(15, OUTPUT);
pinMode(33, OUTPUT);
pinMode(27, OUTPUT);
pinMode(bt1, INPUT_PULLUP);
Serial.begin(9600);
BLEDevice::init("ESP32");
pServer = BLEDevice::createServer();
pServer->setCallbacks(new MyServerCallbacks());
BLEService *pService = pServer->createService(SERVICE_UUID);
pCharacteristic = pService->createCharacteristic(
CHARACTERISTIC_UUID,
BLECharacteristic::PROPERTY_WRITE |
BLECharacteristic::PROPERTY_NOTIFY
);
pCharacteristic->addDescriptor(new BLE2902());
pService->start();
BLEAdvertising *pAdvertising = BLEDevice::getAdvertising();
pAdvertising->addServiceUUID(SERVICE_UUID);
pAdvertising->setScanResponse(false);
pAdvertising->setMinPreferred(0x0);
BLEDevice::startAdvertising();
Serial.println("Waiting a client connection to notify...");
}
void loop()
{
if (bt1g) {
if (digitalRead(bt1) == LOW ) {
bt1t = (bt1t + 1) % 2;
Serial.println(bt1t);
bt1g = false;
}
}
if (!bt1g) {
if (digitalRead(bt1) == HIGH) {
bt1g = true;
}
}
if (bt1t == 0) {
digitalWrite(15, LOW);
digitalWrite(33, HIGH);
}
}
boolean newState = digitalRead(15);
if (deviceConnected) {
if (newState != oldState) {
if (newState == LOW) {
pCharacteristic->setValue("1");
}
else {
pCharacteristic->setValue("0");
}
pCharacteristic->notify();
};
oldState = newState;
}
delay(50);
}
ios 应用的整个代码似乎太长,无法提交到这篇文章,所以here 是 github
我真的不确定并且陷入了困境。任何帮助表示赞赏!
【问题讨论】:
-
我试图查看您的代码,但发现它是多余的。为了从您的 iOS apo 接收写入命令,您不需要循环中的任何代码。当从连接的应用程序发送数据时,将触发写入功能的特征回调。您还试图在循环中操纵这些输出的事实令人困惑。为什么不简化您的示例,删除循环中发生的事情并再次尝试发布。我想帮忙,但在遵循您的逻辑时遇到了麻烦
标签: ios swift arduino bluetooth-lowenergy esp32