基于Xeoncross类的函数

function shortly($input){
$dictionary = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','0','1','2','3','4','5','6','7','8','9'];
if($input===0)
    return $dictionary[0];
$base = count($dictionary);
if(is_numeric($input)){
    $result = [];
    while($input > 0){
        $result[] = $dictionary[($input % $base)];
        $input = floor($input / $base);
    }
    return join("", array_reverse($result));
}
$i = 0;
$input = str_split($input);
foreach($input as $char){
    $pos = array_search($char, $dictionary);
    $i = $i * $base + $pos;
}
return $i;
}

为什么要使用哈希呢?

您只需将自动递增值简单地转换为字母数字值。你可以通过使用一些基数转换很容易做到这一点。假设您的字符空间(a-z, a-z, 0-9等)有62个字符，将id转换为以40为基数的数字并使用这些字符作为数字。

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。

基于Xeoncross类的函数

function shortly($input){
$dictionary = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','0','1','2','3','4','5','6','7','8','9'];
if($input===0)
    return $dictionary[0];
$base = count($dictionary);
if(is_numeric($input)){
    $result = [];
    while($input > 0){
        $result[] = $dictionary[($input % $base)];
        $input = floor($input / $base);
    }
    return join("", array_reverse($result));
}
$i = 0;
$input = str_split($input);
foreach($input as $char){
    $pos = array_search($char, $dictionary);
    $i = $i * $base + $pos;
}
return $i;
}

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)
  }

}

我将继续您的“将数字转换为字符串”方法。但是，如果您的ID是质数且大于52，您将意识到您提出的算法将失败。

理论背景

你需要一个双射函数f。这是必要的，这样你就可以为你的f(123) = 'abc'函数找到一个逆函数g('abc') = 123。这意味着:

一定不存在x1 x2 (x1≠x2)使得f(x1) = f(x2) 对于每一个y，你必须能找到一个x，使f(x) = y。

如何将ID转换为缩短的URL

Think of an alphabet we want to use. In your case, that's [a-zA-Z0-9]. It contains 62 letters. Take an auto-generated, unique numerical key (the auto-incremented id of a MySQL table for example). For this example, I will use 12510 (125 with a base of 10). Now you have to convert 12510 to X62 (base 62). 12510 = 2×621 + 1×620 = [2,1] This requires the use of integer division and modulo. A pseudo-code example: digits = [] while num > 0 remainder = modulo(num, 62) digits.push(remainder) num = divide(num, 62) digits = digits.reverse Now map the indices 2 and 1 to your alphabet. This is how your mapping (with an array for example) could look like: 0 → a 1 → b ... 25 → z ... 52 → 0 61 → 9 With 2 → c and 1 → b, you will receive cb62 as the shortened URL. http://shor.ty/cb

如何将缩短的URL解析为初始ID

反过来就更容易了。你只需要在字母表中反向查找。

E9a62将被解析为“字母表中的第4、61和0个字母”。 E9a62 = [4,61,0] = 4×622 + 61×621 + 0×620 = 1915810 现在找到WHERE id = 19158的数据库记录并执行重定向。

示例实现(由评论者提供)

c++ Python Ruby Haskell c# CoffeeScript Perl