Calculate battery (18650 Li-ion) voltage using ESP8266-12E i am trying to calculate the battery voltage using ESP-12E module. I am using a voltage divider circuit to calculate the same. Here is how my circuit looks like:

And here is how my code looks like:
float volt =0.0;
float temp=0.0;
float r1=320000.0;
float r2=100000.0;

void setup(){
    Serial.begin(115200);
    Serial.println(String(VoltageCalculator()).c_str());
}

float VoltageCalculator(){
  for (int thisReading = 0; thisReading &lt; numReadings; thisReading++) {
    readings[thisReading] = 0;
  }
  Serial.println(analogRead(A0));
  for(int i = 0;i&lt;=10; i++){
    total = total - readings[readIndex];
    // read from the sensor:
    readings[readIndex] = analogRead(A0);
    // add the reading to the total:
    total = total + readings[readIndex];
    // advance to the next position in the array:
    readIndex = readIndex + 1;
  
    // if we're at the end of the array...
    if (readIndex &gt;= numReadings) {
      // ...wrap around to the beginning:
      readIndex = 0;
    }
  
    // calculate the average:
    float analogvalue = total / numReadings;

    //  int analogvalue = analogRead(A0);
    temp = (analogvalue * 3.3) / 1024.0;       
    volt = temp / (r2/(r1+r2));  
  }


  if(DEBUG_ENABLE){
    Serial.println();
    Serial.print(&quot;Actual Voltage : &quot;);
    Serial.print(volt);
    Serial.println();
  }
  
  
  float volt_percent;

  if(volt == 3.1){
    volt_percent = 0;
  } else{
    float volt_diff = volt - 3.1;
    volt_percent = (volt_diff/1.1)*100;
    if(DEBUG_ENABLE){
      Serial.println();
      Serial.print(&quot;Voltage percentage : &quot;);
      Serial.print(volt_percent);
      Serial.println();
    }
  }
  return volt_percent;
}

When i use this circuit with NodeMCU it works very fine, but when it comes to ESP-12E module it do not show the proper voltage. on Serial.println(analogRead(A0)); always shows 1024.
Any suggestions why this is happening and how can i do this to achieve the most accurate value?
Everything was correct and my circuit is also good. The only difference i found is to this below line:
temp = (analogvalue * 3.3) / 1024.0; 
Which should be
temp = (analogvalue * 1.0) / 1024.0;
We are multiplying because we are referencing the voltage from voltage divider circuit as 1 volt.
After doing this, i am getting the exact values battery voltage which then i convert to percentage.
Thank you!