php выдает ошибку Notice: Undefined variable: by in C:\wamp64\www\*\index.php on line 4

Выдает ошибку (!) Notice: Undefined variable: by in C:\wamp64\www\database\manager\index.php on line 4
Вот код (index.php):

<?php
$by = 0;
function enter($d,$t,$c){if($by==0){echo $d.'1'.$t.'2'.$c;}}

$db = new sqlite3("dictionary.db");
if ($results = $db->query("SELECT COUNT(*) FROM translate")) {
    $rows = $results->fetchArray(SQLITE3_ASSOC);
      for ($i=0; $i < $rows['COUNT(*)']; $i++) { 
    $ii = 1+$i;

$engsql = $db->query("SELECT eng FROM users WHERE ID = ".$ii);
$engcom = $engsql->fetchArray(SQLITE3_ASSOC);$eng=preg_replace('/\d/','',$pas);

$russql = $db->query("SELECT rus FROM users WHERE ID = ".$ii);
$ruscom = $russql->fetchArray(SQLITE3_ASSOC);$rus=preg_replace('/\d/','',$log);

$trssql = $db->query("SELECT trs FROM users WHERE ID = ".$ii);
$trscom = $trssql->fetchArray(SQLITE3_ASSOC);$trs=preg_replace('/\d/','',$nick);

$comsql = $db->query("SELECT com FROM users WHERE ID = ".$ii);
$comcom = $comsql->fetchArray(SQLITE3_ASSOC);$com=preg_replace('/\d/','',$name);

    if($_GET['text']==implode('',$eng)){
      enter(implode('',$rus),'['.implode('',$trs).']',implode('',$com));
      $by++;break;
    } else if ($_GET['text']==implode('',$rus)) {
      enter(implode('',$eng),'['.implode('',$trs).']',implode('',$com));
      $by++;
      break;
    } else if ($i<=$rows['COUNT(*)']) {
  enter(' ',' ',' ');
}
  }}

Вопросы, которые уже есть на ru.stackoverflow.com не решили мою проблему