.Net/C#/VB/TSQL/Java 实现: 将天文数字转换成中文大写 (2000 年前的思路,打劫的,一点儿技术含量都没有)

C# Code:
/*
.Net/C#/T-SQL/VB/Java 实现: 将天文数字转换成中文大写
最近这里流行这个大写金额转换
我看了几个,感觉思路都不如这个 2000 年左右用 VB6 写的:
《VB6 之数据格式化对象使用技巧》
http://search.csdn.net/Expert/topic/38/38970.xml?temp=.5078089
<<精华: 将金额小写转大写的代码可转到亿位>>
http://www.dev-club.com/club/bbs/showEssence.asp?id=20684&page=1
思路非常简单,且没有任何位数限制!
例如: 401,0103,1013
读作: 肆佰零壹[亿]零壹佰零叁[万]壹仟零壹拾叁
咱们先按每四位一组 从左到右,高位到低位分别"大声朗读"一下:
"肆佰零壹" 单位是: "[亿]"
"壹佰零叁" 单位是: "[万]"
"壹仟零壹拾叁" 单位是 "" (相当于没有单位)
很容易发现,每四位: 只有 千位,百位,十位,个位 这四种情况!
我们把 [万],[亿] 当作单位就可以了!
这就是规律了!简单吧!
依据该思路,只用区区不到 50 行代码就可以搞定:
只要你能够提供足够多的"单位"
任何天文数字都可以正确转换!
*/
 
namespace Microshaoft
{
    using System;
 
    public class Util
    {
        public static string ConvertNumberToChinese(string x, string[] Nums, string[] Digits, string[] Units)
        {
            string S = ""; //返回值
            int p = 0; //字符位置指针
            int m = x.Length % 4; //取模
 
            // 四位一组得到组数
            int k = (m > 0 ? x.Length / 4 + 1 : x.Length / 4);
 
            // 外层循环在所有组中循环
            // 从左到右 高位到低位 四位一组 逐组处理
            // 每组最后加上一个单位: "[万亿]","[亿]","[万]"
            for (int i = k; i > 0; i--)
            {
                int L = 4;
                if (i == k && m != 0)
                {
                    L = m;
                }
                // 得到一组四位数 最高位组有可能不足四位
                string s = x.Substring(p, L);
                int l = s.Length;
 
                // 内层循环在该组中的每一位数上循环 从左到右 高位到低位
                for (int j = 0; j < l; j++)
                {
                    //处理改组中的每一位数加上所在位: "仟","佰","拾",""(个)
                    int n = Convert.ToInt32(s.Substring(j, 1));
                    if (n == 0)
                    {
                        if (j < l - 1
                            && Convert.ToInt32(s.Substring(j + 1, 1)) > 0 //后一位(右低)
                            && !S.EndsWith(Nums[n]))
                        {
                            S += Nums[n];
                        }
                    }
                    else
                    {
                        //处理 1013 一千零"十三", 1113 一千一百"一十三"
                        if (!(n == 1 && (S.EndsWith(Nums[0]) | S.Length == 0) && j == l - 2))
                        {
                            S += Nums[n];
                        }
                        S += Digits[l - j - 1];
                    }
                }
                p += L;
                // 每组最后加上一个单位: [万],[亿] 等
                if (i < k) //不是最高位的一组
                {
                    if (Convert.ToInt32(s) != 0)
                    {
                        //如果所有 4 位不全是 0 则加上单位 [万],[亿] 等
                        S += Units[i - 1];
                    }
                }
                else
                {
                    //处理最高位的一组,最后必须加上单位
                    S += Units[i - 1];
                }
            }
            return S;
        }
 
        // 测试程序
        private static void Main()
        {
            //数字 数组
            string[] Nums = new string[] {"零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖"};
            //位 数组
            string[] Digits = new string[] {"", "拾", "佰", "仟"};
            //单位 数组
            string[] Units = new string[] {"", "[万]", "[亿]", "[万亿]"};
 
            Console.WriteLine(ConvertNumberToChinese("1100000013", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("2100000000", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("1000000000", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("40000000013", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("40000000001", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("400000010000", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("40101031013", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("40101031113", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("101140101031013", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("100000001000013", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("100000001000113", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("100011003", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("10010103", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("10110013", Nums, Digits, Units));
            Console.WriteLine(ConvertNumberToChinese("130000", Nums, Digits, Units));
            Console.ReadLine();
        }
    }
}

T-SQL UDF Code:
create function ConvertNumberToChinese(@ varchar(100))
returns varchar(100)
as
begin
--declare @ varchar(100)
--set @ = '101140101031013'
declare @s varchar(100)
set @s = ''
 
declare @p integer
set @p = 0
 
declare @m integer
set @m = len(@) % 4
 
declare @k integer
set @k = len(@)/4
 
select @k = @k + 1
 where @m > 0
 
declare @i integer
set @i = @k
 
while (@i > 0)
begin --outer
   declare @L integer
   set @L = 4
 
   select @L = @m
    where @i = @k and @m != 0
 
   declare @ss varchar(4)
   set @ss = substring(@,@p+1,@L)
 
   declare @ll integer
   set @ll = len(@ss)
 
   --inner
   declare @j integer
   set @j = 0
 
   while (@j < @ll) --inner
   begin --inner
      declare @n integer
      set @n = cast(substring(@ss,@j+1,1) as integer)
 
      declare @num varchar(2)   
      select @num = Num
        from (
              select 0 as id,'零' as Num
              union all select 1,'壹'
              union all select 2,'贰'
              union all select 3,'叁'
              union all select 4,'肆'
              union all select 5,'伍'
              union all select 6,'陆'
              union all select 7,'柒'
              union all select 8,'捌'
              union all select 9,'玖'
             ) Nums
      where id = @n
 
      if @n = 0
      begin
         select @s = @s + @num
          where @j < @ll - 1
                and cast(substring(@ss,(@j+1)+1,1) as integer) > 0
                and right(@ss,1) != @num
      end
      else
      begin
         declare @jj integer
         set @jj = 1
 
         select @jj = @j - 1
          where @j > 1
 
         select @s = @s + @num
          where not (@n = 1
                     and @j = @ll - 2
                     and (len(@s) = 0
                          or right(@s,1) = '零'
/*
                                           (   select Num
                                                from
                                               (
                                                select 0 as id,'零' as Num
                                                union all select 1,'壹'
                                                union all select 2,'贰'
                                                union all select 3,'叁'
                                                union all select 4,'肆'
                                                union all select 5,'伍'
                                                union all select 6,'陆'
                                                union all select 7,'柒'
                                                union all select 8,'捌'
                                                union all select 9,'玖'
                                               ) Nums
                                               where id = 0
                                           )
*/
                                           )
                         )
 
         select @s = @s + digit
           from (
                 select 0 as id,'' as digit
                 union all select 1,'拾'
                 union all select 2,'佰'
                 union all select 3,'仟'
                ) digits
          where id = @ll - @j - 1
 
      end
      set @j = @j + 1 --inner
   end --inner
   set @p = @p + @L
 
   declare @unit varchar(10)
   select @unit = Unit
     from (
           select 0 as id,'' as Unit
           union all select 1,'[万]'
           union all select 2,'[亿]'
           union all select 3,'[万亿]'
          ) Units
    where id = @i - 1
 
   if @i < @k
   begin
      select @s = @s + @unit
       where cast(@ss as integer) != 0
   end
   else
   begin
      set @s = @s + @unit
   end
   set @i = @i - 1 -- outer
end --out
return @s
/*
--Test:
select dbo.ConvertNumberToChinese('1011111112101013')
,dbo.ConvertNumberToChinese('40000000001')
,dbo.ConvertNumberToChinese('400000010000')
,dbo.ConvertNumberToChinese('40101031013')
,dbo.ConvertNumberToChinese('101140101031013')
,dbo.ConvertNumberToChinese('100000001000003')
,dbo.ConvertNumberToChinese('10011003')
,dbo.ConvertNumberToChinese('10010103')
,dbo.ConvertNumberToChinese('10010013')
*/
end

VB6 Code:
《VB6 之数据格式化对象使用技巧》
http://search.csdn.net/Expert/topic/38/38970.xml?temp=.5078089
<<精华: 将金额小写转大写的代码可转到亿位>>
http://www.dev-club.com/club/bbs/showEssence.asp?id=20684&page=1

VB.Net Code:
可自行将 C# Code 编译成 EXE 后, Reflector 反编译获取 VB.Net/Delphi.Net/IL 等语言源码

Java Code:
public class Class1
{
    public static String ConvertNumberToChinese(String x, String[] Nums, String[] Digits, String[] Units)
    {
        String S = ""; //返回值
        int p = 0; //字符位置指针
        int m = x.length() % 4; //取模
 
        // 四位一组得到组数
        int k = (m > 0 ? x.length() / 4 + 1 : x.length() / 4);
 
        // 外层循环在所有组中循环
        // 从左到右 高位到低位 四位一组 逐组处理
        // 每组最后加上一个单位: "[万亿]","[亿]","[万]"
        for (int i = k; i > 0; i--)
        {
            int L = 4;
            if (i == k && m != 0)
            {
                L = m;
            }
            // 得到一组四位数 最高位组有可能不足四位
            String s = x.substring(p, p + L);
            int l = s.length();
 
            // 内层循环在该组中的每一位数上循环 从左到右 高位到低位
            for (int j = 0; j < l; j++)
            {
                //处理改组中的每一位数加上所在位: "仟","佰","拾",""(个)
                int n = java.lang.Integer.parseInt(s.substring(j, j+1));
                if (n == 0)
                {
                    if ((j < l - 1)
                        && (java.lang.Integer.parseInt(s.substring(j + 1, j + 1+ 1)) > 0) //后一位(右低)
                        && !S.endsWith(Nums[n]))
                    {
                        S += Nums[n];
                    }
                }
                else
                {
                    //处理 1013 一千零"十三", 1113 一千一百"一十三"
                    if (!(n == 1 && (S.endsWith(Nums[0]) | S.length() == 0) && j == l - 2))
                    {
                        S += Nums[n];
                    }
                    S +=  Digits[l - j - 1];
                }
            }
            p += L;
            // 每组最后加上一个单位: [万],[亿] 等
            if (i < k) //不是最高位的一组
            {
                if (java.lang.Integer.parseInt(s) != 0)
                {
                    //如果所有 4 位不全是 0 则加上单位 [万],[亿] 等
                    S += Units[i - 1];
                }
            }
            else
            {
                //处理最高位的一组,最后必须加上单位
                S += Units[i - 1];
            }
        }
        return S;
    }
 
    // 测试程序
    public static void main(String[] args) throws Exception
    {
        //数字 数组
        String[] Nums = new String[] {"零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖"};
        //位 数组
        String[] Digits = new String[] {"", "拾", "佰", "仟"};
        //单位 数组
        String[] Units = new String[] {"", "[万]", "[亿]", "[万亿]"};
 
        System.out.println(ConvertNumberToChinese("111112100113", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("1100000000", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("1000000000", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("40000000013", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("40000000001", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("400000010000", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("40101031013", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("40101031113", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("101140101031013", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("100000001000013", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("100000001000113", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("100011003", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("10010103", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("10110013", Nums, Digits, Units));
        System.out.println(ConvertNumberToChinese("130000", Nums, Digits, Units));
 
        //System.in.read();
    }
}