html强制转换为数字,如何在HTML中将基数为10的数字转换为字母顺序列表

我想将整数转换为字母等价,如

HTML中的有序列表.

我试图将基数为10的数字转换为带有-z数字的基数26.

但那不是我想要的.

IN WANT GET

-----------------------

1 => a <= a

2 => b <= b

3 => c <= c

4 => d <= d

5 => e <= e

6 => f <= f

7 => g <= g

8 => h <= h

9 => i <= i

10 => j <= j

11 => k <= k

12 => l <= l

13 => m <= m

14 => n <= n

15 => o <= o

16 => p <= p

17 => q <= q

18 => r <= r

19 => s <= s

20 => t <= t

21 => u <= u

22 => v <= v

23 => w <= w

24 => x <= x

25 => y <= y

26 => z <= az

27 => aa <= aa

28 => ab <= ab

29 => ac <= ac

private final static char[] digits = {

'0','a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'

};

private static String numberToAlphaNumeric(long i,int radix) {

char[] buf = new char[65];

int charPos = 64;

boolean negative = (i < 0);

if (!negative) {

i = -i;

}

while (i <= -radix) {

buf[charPos--] = digits[(int)(-(i % radix))];

i = i / radix;

}

buf[charPos] = digits[(int)(-i)];

if (negative) {

buf[--charPos] = '-';

}

return new String(buf,charPos,(65 - charPos));

}

public static String numberToAlphaNumeric(long number) {

ArrayList list = new ArrayList();

for( int j = 0; list.size() != number; j++ ) {

String alpha = numberToAlphaNumeric( j,digits.length );

if(!alpha.contains( "0" )) {

list.add( alpha );

}

}

return list.get( list.size()-1 );

}

我的第二个想法：

如果我将新的前导符号扩展为数字并将我的数字转换为基数为27的数字,

我在每次携带中都有新的符号,这是错误的,我可以过滤掉它们.

这是非常低效和丑陋的,但我没有更多的想法.常见的方法是什么？