Question

我想每个人都会认为这很奇怪，实际上它确实很奇怪。我一直困扰并尝试一切可能的方法来解决这个问题。问题是，如果我在薄膜键盘上按键后删除执行方法的逻辑，则会成功执行HTTP post方法。它不与服务器连接。但是一旦我按下按键并将逻辑放入流程后删除执行，它将产生一个成功的post方法以下是代码。非常感谢任何帮助。

//Include libraries
#include <SPI.h>
#include <WiFi.h>
#include <Keypad.h>
#include <LiquidCrystal.h>

const byte ROWS = 4;
const byte COLS = 4;
char keys[ROWS][COLS] = {
  {'1', '2', '3', 'A'},
  {'4', '5', '6', 'B'},
  {'7', '8', '9', 'C'},
  {'*', '0', '#', 'D'}
};
byte rowPins[ROWS] = {52, 26, 48, 46};
byte colPins[COLS] = {42, 40, 38, 36};
Keypad keypad = Keypad( makeKeymap(keys), rowPins, colPins, ROWS, COLS );
LiquidCrystal lcd(45, 43, 37, 35, 33, 31);
WiFiClient client;
// EDIT: 'Server' address to match your domain
char server[] = "https://e-medic.herokuapp.com";

#include <system.h>

void setup() {
  Serial.begin(9600);
  //wifi connect (I have not put the method but wifi will be connected)
  connectWifi();
  printWifiStatus();
  //lcd start
  lcd.begin(16, 2);
  lcdPrint(0);
  initialState();
}

void loop() {
  //checking whether the client sends some data back
  while (client.available()) {
    char c = client.read();
    Serial.write(c);
  }
}

void initialState(char key) {
  char key = keypad.getKey();
  while (key != 'A') {
    Serial.println("you havent press A");
    key = keypad.getKey();
  }
  Serial.println("Start of Post");
  // If there's a successful connection, send the HTTP POST request
  value1 = "test";
  //Appending the data to a variable to be posted.
  String http_content = "temp=" + value1;
  Serial.println("before connecting to server...");
  //checking whether the client is connected
  //client will not make connection with server but once I remove the
  //keypad press it will connect to server
  if (client.connect(server, 80)) {
    Serial.println("after...");
    Serial.println("connecting. now inside connected.");
    client.println("POST /api/uploadPatientData HTTP/1.1");
    // EDIT: 'Host' to match your domain
    client.println("Host: e-medic.herokuapp.com");
    client.println("User-Agent: Arduino/1.0");
    client.println("Connection: close");
    client.println("Content-Type: application/x-www-form-urlencoded;");
    client.print("Content-Length: ");
    client.println(http_content.length());
    Serial.println(http_content.length());
    client.println();
    client.println(http_content);
    delay(5000);
  } else {
    // If you couldn't make a connection:
    Serial.println("Connection failed");
    Serial.println("Disconnecting.");
    client.stop();
  }
  char c = client.read();
  Serial.write(c);
}

在薄膜键盘按键上调用时无法执行HTTP post方法

0 个答案: