Repetitive 0's?

so posted topic few days ago , couple people helped me on how input multiple digit variables keypad. followed advice , can see 0's now. come in groups of 3 , spam lcd screen. would've posted in other topic didn't want confusion title. being said, hope else has experienced , solved or sees error in code.

i have code below , attach photo of problem.

code: [select]



/* code allows top input values via 3x4 keypad display
on lcd screen in number format , alter rgb led color. can switch
which color edit * button. use # update colours
pins:
arduino lcd interface
5v vcc
gnd gnd
a5 scl
a4 sda

arduino keypad
8 1
7 2
6 3
5 4
4 5
3 6
2 7

arduino rgb
~220ohm resistors between digital pins , color pins
9 red
10 blue
11 green
gnd gnd
*/

#include <wire.h>
#include <liquidcrystal_i2c.h>
liquidcrystal_i2c lcd(0x27, 2, 1, 0, 4, 5, 6, 7, 3, positive);

#include <keypad.h>

int yr = 0;
int yg = 0;
int yb = 0;


int digitra = 0;
int digitrb = 0;
int digitrc = 0;
int reddigit = 1;


int digitga= 0;
int digitgb= 0;
int digitgc= 0;
int greendigit = 1;

int digitba = 0;
int digitbb = 0;
int digitbc = 0;
int bluedigit = 1;

void keyred(int x){
  if (reddigit==1){
    digitra = x;
    lcd.print(x);
  }
  if (reddigit==2){
    digitrb = x;
    lcd.print(x);
  }
  if (reddigit==3){
    digitrc = x;
    lcd.print(x);
  }
  reddigit = reddigit + 1;
  if (reddigit >=4){}
}


void keygreen(int x){
  if (reddigit==1){
    digitga = x;
    lcd.print(x);
  }
  if (reddigit==2){
    digitgb = x;
    lcd.print(x);
  }
  if (reddigit==3){
    digitgc = x;
    lcd.print(x);
  }
  greendigit = greendigit + 1;
  if (greendigit >=4){}
}


void keyblue(int x){
  if (bluedigit==1){
    digitba = x;
    lcd.print(x);
  }
  if (reddigit==2){
    digitbb = x;
    lcd.print(x);
  }
  if (reddigit==3){
    digitbc = x;
    lcd.print(x);
  }
  bluedigit = bluedigit + 1;
  if (bluedigit >=4){}
}

int transform(int a, int b, int c){
  int x;
  x = a*100 + b*10 + c;
  return x;
}
  
  
int color = 1;  //allows control on color writing

int redpin = 9;
int greenpin = 10;  //sets individual color pins
int bluepin =11;



const byte rows = 4; //four rows
const byte cols = 3; //three columns
char keys[rows][cols] = {
  {'1','2','3'},
  {'4','5','6'}, //creates keypad
  {'7','8','9'},
  {'*','0','#'}
};
byte rowpins[rows] = {5, 4, 3, 2}; //connect row pinouts of keypad
byte colpins[cols] = {8, 7, 6}; //connect column pinouts of keypad


keypad keypad = keypad( makekeymap(keys), rowpins, colpins, rows, cols );

void setup(){
  lcd.begin(16,2);
  lcd.clear();
  
  //test
  serial.begin(9600);
  //test

}

void loop(){
  char key = keypad.getkey();
  
  if(key == '#'){
   lcd.clear();
   analogwrite(redpin, transform(digitra, digitrb, digitbc));
   analogwrite(greenpin, transform(digitga, digitgb, digitgc));
   analogwrite(bluepin, transform(digitba, digitbb, digitbc));
   lcd.setcursor(0,1);
   lcd.write("r");
   lcd.print(transform(digitra, digitrb, digitbc));
   lcd.write("g");
   lcd.print(transform(digitga, digitgb, digitgc));
   lcd.write("b");
   lcd.print(transform(digitba, digitbb, digitbc));
  }
  
  if(key == '*'){
  color+=1;
  }
  if(color >= 4){
   color = 1;
  }
  if(key >=0){
  
    if(color == 1){
      keyred(key);
    }
    
    if(color == 2){
      keygreen(key);
    }
    
    if(color == 3){
      keyblue(key);
    }

  }  
}


there few comments here , there have given trying put them in when began redo of code. always, new. saying stuff beyond basics hard pressed me understand.

i start making change loop() follows:

code: [select]


void loop(){
  char key = keypad.getkey();

  if (key != no_key){         // add new line
     serial.print("key = ");  // add new line
     serial.println(key);     // add new line

     // put of existing code here
 
  }                           // add new line
}



as code stands now, execute of code when there no key press. means burning though code hundreds, if not thousands, of times each second, throwing zeroes on screen. run code above , pay attention values see. shouldn't see until press key. give starting place what's happening.

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