create or replace function F_upper_money(p_num in number default null)
  return nvarchar2 is
  Result      nvarchar2(100); --返回字符串
  num_round   nvarchar2(100) := to_char(abs(round(p_num, 2))); --转换数字为小数点后2位的字符(正数)
  num_left    nvarchar2(100); --小数点左边的数字
  num_right   nvarchar2(2); --小数点右边的数字
  str1        nchar(10) := '零壹贰参肆伍陆柒捌玖'; --数字大写
  str2        nchar(16) := '元拾佰仟万拾佰仟亿拾佰仟万拾佰仟'; --数字位数(从低至高)
  num_pre     number(1) := 1; --前一位上的数字
  num_current number(1); --当前位上的数字
  num_count   number := 0; --当前数字位数

begin
  if p_num is null then
    return null;
  end if; --转换数字为null时返回null

  select to_char(nvl(substr(to_char(num_round),
                            1,
                            decode(instr(to_char(num_round), '.'),
                                   0,
                                   length(num_round),
                                   instr(to_char(num_round), '.') - 1)),
                     0))
    into num_left
    from dual; --取得小数点左边的数字
  select substr(to_char(num_round),
                decode(instr(to_char(num_round), '.'),
                       0,
                       length(num_round) + 1,
                       instr(to_char(num_round), '.') + 1),
                2)
    into num_right
    from dual; --取得小数点右边的数字

  if length(num_left) > 16 then
    return '**********';
  end if; --数字整数部分超过16位时

  --采用从低至高的算法，先处理小数点右边的数字
  if length(num_right) = 2 then
    if to_number(substr(num_right, 1, 1)) = 0 then
      result := '零' ||
                substr(str1, to_number(substr(num_right, 2, 1)) + 1, 1) || '分';
    else
      result := substr(str1, to_number(substr(num_right, 1, 1)) + 1, 1) || '角' ||
                substr(str1, to_number(substr(num_right, 2, 1)) + 1, 1) || '分';
    end if;
  elsif length(num_right) = 1 then
    result := substr(str1, to_number(substr(num_right, 1, 1)) + 1, 1) || '角整';
  else
    result := '整';
  end if;
  --再处理小数点左边的数字
  for i in reverse 1 .. length(num_left) loop
    --(从低至高)
    num_count   := num_count + 1; --当前数字位数
    num_current := to_number(substr(num_left, i, 1)); --当前位上的数字
    if num_current > 0 then
      --当前位上数字不为0按正常处理
      result := substr(str1, num_current + 1, 1) ||
                substr(str2, num_count, 1) || result;
    else
      --当前位上数字为0时
      if mod(num_count - 1, 4) = 0 then
        --当前位是元、万或亿时
        result  := substr(str2, num_count, 1) || result;
        num_pre := 0; --元、万,亿前不准加零
      end if;
      if num_pre > 0 or length(num_left) = 1 then
        --上一位数字不为0或只有个位时
        result := substr(str1, num_current + 1, 1) || result;
      end if;
    end if;
    num_pre := num_current;
  end loop;

  if p_num < 0 then
    --转换数字是负数时
    result := '负' || result;
  end if;

  return Result;

exception
  when others then
    raise_application_error(-20001, '数字转换大写出现错误！' || sqlerrm);
end;