由于 php 中的大量数据库请求导致服务器错误 [关闭]

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here is the code for it please solve it and here is the error screenshottt

Error 524

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$sql = "SELECT * FROM `users`";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_assoc($result)) {
  $uid = $row['uid'];
  {
    $_SESSION['referralIncome'] = $uid;
    $uidglb = $_SESSION['referralIncome'];
    $alldirects = array($uid);
    $alldirectstemp = [];
    $sno = 1;
    $level = 1;
    $levels = [];
    $sizeofalldirects = 0;
    $referralIncome = 0;
    $perReferralIncome = 5;
    referralIncome();
  }
}

function referralIncome()
{
  global $alldirects, $alldirectstemp, $conn, $sno, $level, $levels, $sizeofalldirects, $perReferralIncome, $uidglb;
  foreach ($alldirects as $direct) {
    $sql = "SELECT * FROM `users` WHERE `sponsorid`='$direct' and `plan`='1' ORDER BY `datetime`";
    $result = mysqli_query($conn, $sql);
    array_shift($alldirects);
    while ($row = mysqli_fetch_assoc($result)) {
      array_push($alldirectstemp, $row['uid']);
      $rowuid = $row['uid'];
      $plan = mysqli_fetch_assoc(mysqli_query($conn, "SELECT `amount` from `plan` WHERE `uid`='$rowuid'"))['amount'];
      $perReferralIncome;
      if ($perReferralIncome < 0) {
        $perReferralIncome = 0;
      }
      $incomeReferral = ($perReferralIncome * 0.01) * $plan;

      function2referralIncome($row, $incomeReferral);
      $sno++;
      $sizeofalldirects++;
    }
  }
  $alldirects = $alldirectstemp;
  if (!$alldirects == array()) {
    $perReferralIncome--;
    $level++;
    array_push($levels, $sizeofalldirects);
    $sizeofalldirects = 0;
    $alldirectstemp = array();
    $sql = "SELECT `uid` FROM `levels` WHERE `uid`='$uidglb'";
    $isLevelsAdded = mysqli_num_rows(mysqli_query($conn, $sql));
    if (!$isLevelsAdded > 0) {
      $sql = "INSERT INTO `levels` (`uid`, `1`, `2`, `3`, `4`, `5`) VALUES ('$uidglb','" . $levels[0] ?? 0 . "','" . $levels[1] ?? 0 . "','" . $levels[2] ?? 0 . "','" . $levels[3] ?? 0 . "','" . $levels[4] ?? 0 . "',)";
    } else {
      $sql = "UPDATE `levels` SET `1`='" . $levels[0] ?? 0 . "',`2`='" . $levels[1] ?? 0 . "',`3`='" . $levels[2] ?? 0 . "',`4`='" . $levels[3] ?? 0 . "',`5`='" . $levels[4] ?? 0 . "' WHERE `uid`='$uidglb'";
    }
    $result = mysqli_query($conn, $sql);
    referralIncome();
  }
}