Arduino IDE/ESP8266：为什么必须按两次按钮才能获得正确的输出？

Question

我一直在自学 ESP8226 编程，特别是 ESP8226-12F。我有这个基于在线示例的脚本：

#include <ESP8266WiFi.h>

WiFiServer server(80); //Initialize the server on Port 80

void setup() {
  WiFi.mode(WIFI_AP); //Our ESP8266-12E is an AccessPoint
  WiFi.begin("***","***"); // Provide the (SSID, password);
  Serial.begin(115200); //Start communication between the ESP8266-12E and the monitor window
  IPAddress HTTPS_ServerIP= WiFi.localIP(); // Obtain the IP of the Server
  Serial.print("Server IP is: "); // Print the IP to the monitor window
  Serial.println(HTTPS_ServerIP);
  while(WiFi.status() != WL_CONNECTED) {
    delay(500);
    Serial.print(".");
  }

  Serial.println(WiFi.localIP());
  server.begin(); // Start the HTTP Server
}

void loop() {
  String status;
  WiFiClient client = server.available();
  if (!client) {
    return;
  }
  Serial.println("Somebody has connected :)");
  //Read what the browser has sent into a String class and print the request to the monitor
  String request = client.readString();
  Serial.println(request);
  // Handle the Request
  if (request.indexOf("/CLOSE") != -1) {
    Serial.println(request.indexOf("/CLOSE"));
    status = "CLOSED";
  }
  if (request.indexOf("/OPEN") != -1) {
    Serial.println(request.indexOf("/OPEN"));
    status = "OPEN";
  }
  Serial.println(status);
  // The HTML
  String s = "HTTP/1.1 200 OK";
  s += "Content-Type: text/html\r\n\r\n";
  s += "<!DOCTYPE html><html><head><meta http-equiv=\"Content-Security-Policy\"   content=\"default-src 'self'\"> <meta name=\"referrer\" content=\"no-referrer\" /></head><body>";
  s += "<br><input type=\"button\" name=\"b1\" value=\"OPEN DOOR\" onclick=\"javascript:location.href='/OPEN'\">";
  s += "<br><input type=\"button\" name=\"b2\" value=\"CLOSE DOOR\" onclick=\"javascript:location.href='/CLOSE'\">";
  if(status == "OPEN") {
    s+= "<br><br><p>OPEN</p>";
  } else {
    s+= "<br><br><p>CLOSED</p>";
  }
  s += "</body></html>\n";
  client.flush(); //clear previous info in the stream
  client.print(s); // Send the response to the client
  delay(1);
  Serial.println("Client disonnected");

您会看到初始 HTTP 响应创建了两个按钮：OPEN DOOR 和 CLOSE DOOR。

我遇到的问题是必须按两次关闭按钮才能使草图输出 "CLOSED"。

这是串行监视器的输出。

1: 第一次浏览到 http://10.0.1.92:

Somebody has connected :)
GET / HTTP/1.1
Host: 10.0.1.92
User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.12; rv:5.0.1) Gecko/20100101 Firefox/5.0.1
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-us,en;q=0.5
Accept-Encoding: gzip, deflate
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7
Connection: keep-alive

Client disonnected

2: 按打开按钮 http://10.0.1.92/OPEN 显示 "OPEN"

Somebody has connected :)
GET /OPEN HTTP/1.1
Host: 10.0.1.92
User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.12; rv:5.0.1) Gecko/20100101 Firefox/5.0.1
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-us,en;q=0.5
Accept-Encoding: gzip, deflate
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7
Connection: keep-alive
Referer: http://10.0.1.92/


4  // This is value of request.indexOf("/CLOSE")
OPEN
Client disonnected

3: 按CLOSE http://10.0.1.92/CLOSE 仍然显示"OPEN"

Somebody has connected :)
GET /CLOSE HTTP/1.1
Host: 10.0.1.92
User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.12; rv:5.0.1) Gecko/20100101 Firefox/5.0.1
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-us,en;q=0.5
Accept-Encoding: gzip, deflate
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7
Connection: keep-alive
Referer: http://10.0.1.92/OPEN


4 
371 // No idea where this comes from
OPEN
Client disonnected

4：第二次按CLOSE，显示"CLOSED"：

Somebody has connected :)
GET /CLOSE HTTP/1.1
Host: 10.0.1.92
User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.12; rv:5.0.1) Gecko/20100101 Firefox/5.0.1
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-us,en;q=0.5
Accept-Encoding: gzip, deflate
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7
Connection: keep-alive
Referer: http://10.0.1.92/CLOSE


4
CLOSED
Client disonnected

为什么要按两次关闭按钮才能显示 "CLOSED"？这是逻辑问题，还是 HTTP 问题，还是两者的某种结合？

Answer 1

这是可疑的：

if(status == "OPEN"){
   s+= "<br><br><p>"+status+"</p>";
} else {
   s+= "<br><br><p>"+status+"</p>";
}

如果两个选项相同，if 语句是什么？

另外 Referer 显示了三个不同的值，而根据您的设计，一个人会期望两个值。我想知道你最初在哪里设置这个二进制状态机的状态？

编辑：

所以 OP 更新了源代码。

if(status == "OPEN") {
    s+= "<br><br><p>OPEN</p>";
  } else {
    s+= "<br><br><p>CLOSED</p>";
  }

对战：

 if (request.indexOf("/CLOSE") != -1) {
    Serial.println(request.indexOf("/CLOSE"));
    status = "CLOSED";
  }
  if (request.indexOf("/OPEN") != -1) {
    Serial.println(request.indexOf("/OPEN"));
    status = "OPEN";
  }

你现在看到它们在逻辑上有何不同了吗？

这可能意味着当状态机启动时，由于它没有打开，所以它在顶部块中关闭但在底部块中没有关闭。按下下一个按钮将其更改为打开，然后第二次将其更改为关闭？这可能是正在发生的事情吗？

Answer 2

问题是这些语句：

if (request.indexOf("/CLOSE") != -1){  . . .
if (request.indexOf("/OPEN") != -1){  . . .

如果您查看响应字符串，您会发现那些 if 语句与 GET /OPEN 或 GET /CLOSED 以及 Referer 相匹配：http://10.0.1.92/CLOSE and Referer: http://10.0.1.92/OPEN.

那个371一开始没看懂是Referer中/OPEN的字符串位置：http://10.0.1.92/OPEN。该匹配项将状态翻转回 OPEN，直到 Referer 更改并产生观察到的行为。

我将 if 语句更改为包含 space，这样就只会匹配 GET /CLOSE 或 GET /OPEN，即

if (request.indexOf(" /CLOSE") != -1){  . . .
if (request.indexOf(" /OPEN") != -1){  . . .

并且它可以正常工作。所以，逻辑上没有任何问题，它正在做它应该做的事情 ;)。

感谢 JLH 的帮助。

Arduino IDE/ESP8266：为什么必须按两次按钮才能获得正确的输出？

Arduino IDE/ESP8266: Why does a button have to be pressed twice to get the correct output?

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