我有一个问题的变体，因为我存储了来自许多不同作者的网页，需要防止通过猜测发现页面。所以我的短url在Base-62字符串中添加了几个额外的数字作为页码。这些额外的数字是由页面记录本身的信息生成的，它们确保3844个url中只有一个是有效的(假设2位Base-62)。您可以在http://mgscan.com/MBWL上看到概要描述。

非常好的回答，我已经创建了一个bjf的Golang实现:

package bjf

import (
    "math"
    "strings"
    "strconv"
)

const alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"

func Encode(num string) string {
    n, _ := strconv.ParseUint(num, 10, 64)
    t := make([]byte, 0)

    /* Special case */
    if n == 0 {
        return string(alphabet[0])
    }

    /* Map */
    for n > 0 {
        r := n % uint64(len(alphabet))
        t = append(t, alphabet[r])
        n = n / uint64(len(alphabet))
    }

    /* Reverse */
    for i, j := 0, len(t) - 1; i < j; i, j = i + 1, j - 1 {
        t[i], t[j] = t[j], t[i]
    }

    return string(t)
}

func Decode(token string) int {
    r := int(0)
    p := float64(len(token)) - 1

    for i := 0; i < len(token); i++ {
        r += strings.Index(alphabet, string(token[i])) * int(math.Pow(float64(len(alphabet)), p))
        p--
    }

    return r
}

托管在github: https://github.com/xor-gate/go-bjf

这是我的PHP 5类。

<?php
class Bijective
{
    public $dictionary = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";

    public function __construct()
    {
        $this->dictionary = str_split($this->dictionary);
    }

    public function encode($i)
    {
        if ($i == 0)
        return $this->dictionary[0];

        $result = '';
        $base = count($this->dictionary);

        while ($i > 0)
        {
            $result[] = $this->dictionary[($i % $base)];
            $i = floor($i / $base);
        }

        $result = array_reverse($result);

        return join("", $result);
    }

    public function decode($input)
    {
        $i = 0;
        $base = count($this->dictionary);

        $input = str_split($input);

        foreach($input as $char)
        {
            $pos = array_search($char, $this->dictionary);

            $i = $i * $base + $pos;
        }

        return $i;
    }
}

我的Python 3版本

base_list = list("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
base = len(base_list)

def encode(num: int):
    result = []
    if num == 0:
        result.append(base_list[0])

    while num > 0:
        result.append(base_list[num % base])
        num //= base

    print("".join(reversed(result)))

def decode(code: str):
    num = 0
    code_list = list(code)
    for index, code in enumerate(reversed(code_list)):
        num += base_list.index(code) * base ** index
    print(num)

if __name__ == '__main__':
    encode(341413134141)
    decode("60FoItT")

Scala中的实现:

class Encoder(alphabet: String) extends (Long => String) {

  val Base = alphabet.size

  override def apply(number: Long) = {
    def encode(current: Long): List[Int] = {
      if (current == 0) Nil
      else (current % Base).toInt :: encode(current / Base)
    }
    encode(number).reverse
      .map(current => alphabet.charAt(current)).mkString
  }
}

class Decoder(alphabet: String) extends (String => Long) {

  val Base = alphabet.size

  override def apply(string: String) = {
    def decode(current: Long, encodedPart: String): Long = {
      if (encodedPart.size == 0) current
      else decode(current * Base + alphabet.indexOf(encodedPart.head),encodedPart.tail)
    }
    decode(0,string)
  }
}

使用Scala测试的测试示例:

import org.scalatest.{FlatSpec, Matchers}

class DecoderAndEncoderTest extends FlatSpec with Matchers {

  val Alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"

  "A number with base 10" should "be correctly encoded into base 62 string" in {
    val encoder = new Encoder(Alphabet)
    encoder(127) should be ("cd")
    encoder(543513414) should be ("KWGPy")
  }

  "A base 62 string" should "be correctly decoded into a number with base 10" in {
    val decoder = new Decoder(Alphabet)
    decoder("cd") should be (127)
    decoder("KWGPy") should be (543513414)
  }

}

为什么不直接将id转换为字符串呢?您只需要一个函数将0到61之间的数字映射到单个字母(大写/小写)或数字。然后应用它来创建，比如说，4个字母的代码，你就有了1470万个url。