Blynk.syncVirtual(V1)未更新虚拟管脚值
Blynk.syncVirtual(V1)未更新虚拟管脚值
提问于 2020-05-29 19:01:42
我正在尝试这段代码来运行步进电机并打印Blynk应用程序操纵杆的X和Y坐标。但是代码只获取一次操纵杆的值。但是当我使用if条件而不是while()时，它工作得很好。我需要while条件来连续运行步进电机，但在if条件下，它们会非常迅速地关闭和开启，从而降低了步进电机的速度。
请帮我处理这种情况
代码
#define BLYNK_PRINT Serial
#include <ESP8266WiFi.h>
#include <BlynkSimpleEsp8266.h>
#include <AccelStepper.h>
// You should get Auth Token in the Blynk App.
// Go to the Project Settings (nut icon).
char auth[] = "LRTCZUnCI06P-pqh5rlPXRbuOUgQ_uGH";
AccelStepper stepper_1(1,D2,D1);
AccelStepper stepper_2(1,D5,D6);
// Your WiFi credentials.
// Set password to "" for open networks.
char ssid[] = "Airtel_7599998800";
char pass[] = "air71454";
int prev_x = 0;
int prev_y = 0;
BLYNK_WRITE(V1) {
int x = param[0].asInt();
int y = param[1].asInt();
// Do something with x and y
Serial.print("X = ");
Serial.print(x);
Serial.print("; Y = ");
Serial.println(y);
MoveControls(x , y);
}
void setup()
{
// Debug console
Serial.begin(9600);
stepper_1.setAcceleration(1000);
stepper_2.setAcceleration(1000);
Blynk.begin(auth, ssid, pass);
// You can also specify server:
//Blynk.begin(auth, ssid, pass, "blynk-cloud.com", 80);
//Blynk.begin(auth, ssid, pass, IPAddress(192,168,1,100), 8080);
// ESP.wdtDisable();
// ESP.wdtEnable(WDTO_8S);
}
void loop()
{// ESP.wdtFeed();
Blynk.run();
// Serial.println("A");
}
void MoveControls(int x, int y) {
///////////////////////////////////////////Move Forward////////////////////////////////////////////////////
if(y >= 150 && x <= 150 && x >= -150){
stepper_1.enableOutputs();
stepper_2.enableOutputs();
stepper_1.setMaxSpeed(1000);
stepper_2.setMaxSpeed(1000);
stepper_1.move(1000);
stepper_2.move(1000);
while(x != prev_x && y != prev_y){
stepper_1.run();
stepper_2.run();
Blynk.syncVirtual(V1);
Blynk.run();
Serial.print("X = ");
Serial.print(x);
Serial.print("; Y = ");
Serial.println(y);
}
prev_x = x;
prev_y = y;
}
///////////////////////////////////////////Neutral Zone////////////////////////////////////////////////////
if(y <= 150 && x <= 150 && x >= -150 && y >= -150){
stepper_1.disableOutputs();
stepper_2.disableOutputs();
prev_x = x;
prev_y = y;
}
yield();
}这就是while循环
while(x != prev_x && y != prev_y){
stepper_1.run();
stepper_2.run();
Blynk.syncVirtual(V1);
Blynk.run();
Serial.print("X = ");
Serial.print(x);
Serial.print("; Y = ");
Serial.println(y);
}
prev_x = x;
prev_y = y;
}我知道代码非常错误，没有多大意义，但我只需要修复Blynk.syncVirtual()问题。
我还尝试在Blynk.syncVirtual()之后添加Blynk.run()，因为在Blynk Community中有人告诉我要这样做https://community.blynk.cc/t/blynk-syncvirtual-doesnt-work-as-expected/40047/4
回答 2
Stack Overflow用户
发布于 2020-05-29 19:14:56
在while循环中，设置了prev_x = x && prev_y = y，它满足while循环的条件
Stack Overflow用户
发布于 2020-05-29 19:31:32
这行得通吗？
while(y <= 150 && x <= 150 && x >= -150 && y >= -150) {
stepper_1.run();
stepper_2.run();
Blynk.syncVirtual(V1);
Blynk.run();
Serial.print("X = ");
Serial.print(x);
Serial.print("; Y = ");
Serial.println(y);
}
//prev_x = x;
//prev_y = y;页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/62084409
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