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“关闭”。
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:按下打开按钮 http://10.0.1.92/OPEN并显示“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: 按关闭 http://10.0.1.92/CLOSE并且仍然显示“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 CLOSE 按钮必须按两次才能显示“CLOSED”？这是逻辑问题，还是 HTTP 问题，还是两者的某种组合？
最佳答案
这是可疑的:
if(status == "OPEN"){
s+= "<br><br><p>"+status+"</p>";
} else {
s+= "<br><br><p>"+status+"</p>";
}
如果两个备选方案相同，if 语句是什么？
此外，Referer 显示三个不同的值，而根据您的设计，一个人会期望两个值。我想知道您最初在哪里设置这个二进制状态机的状态？
编辑:
所以 OP 已经更新了源代码。
if(status == "OPEN") {
s+= "<br><br><p>OPEN</p>";
} else {
s+= "<br><br><p>CLOSED</p>";
}
对比:
if (request.indexOf("/CLOSE") != -1) {
Serial.println(request.indexOf("/CLOSE"));
status = "CLOSED";
}
if (request.indexOf("/OPEN") != -1) {
Serial.println(request.indexOf("/OPEN"));
status = "OPEN";
}
您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 block 中关闭但在底部 block 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？
关于http - Arduino 集成开发环境/ESP8266 : Why does a button have to be pressed twice to get the correct output?，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44655184/