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;
}

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;
    }
}

您可以散列整个URL，但如果您只想缩短id，请按照marcel的建议进行。我写了这个Python实现:

https://gist.github.com/778542

这是我最初的想法，可以做更多的思考，或者做一些模拟，看看是否有效或需要改进:

我的答案是记住数据库中的长URL，并使用ID 0到999999999999999999(或所需的任意大的数字)。

但是ID 0到999999999999999999可能是个问题，因为

如果我们使用十六进制，甚至base62或base64，它可以更短。(base64就像YouTube使用A-Z A-Z 0-9 _和-) 如果它从0均匀地增加到999999999999999999，那么黑客就可以按照这个顺序访问它们，并知道人们相互发送的url，所以这可能是一个隐私问题

我们可以这样做:

have one server allocate 0 to 999 to one server, Server A, so now Server A has 1000 of such IDs. So if there are 20 or 200 servers constantly wanting new IDs, it doesn't have to keep asking for each new ID, but rather asking once for 1000 IDs for the ID 1, for example, reverse the bits. So 000...00000001 becomes 10000...000, so that when converted to base64, it will be non-uniformly increasing IDs each time. use XOR to flip the bits for the final IDs. For example, XOR with 0xD5AA96...2373 (like a secret key), and the some bits will be flipped. (whenever the secret key has the 1 bit on, it will flip the bit of the ID). This will make the IDs even harder to guess and appear more random

按照这种方案，分配id的单个服务器可以组成id，请求分配id的20或200个服务器也可以组成id。分配服务器必须使用锁/信号量来防止两个请求服务器获得相同的批处理(或者如果它一次接受一个连接，这已经解决了问题)。因此，我们不希望等待分配的队列太长。所以这就是为什么一次分配1000或10000个可以解决问题。

alphabet = map(chr, range(97,123)+range(65,91)) + map(str,range(0,10))

def lookup(k, a=alphabet):
    if type(k) == int:
        return a[k]
    elif type(k) == str:
        return a.index(k)


def encode(i, a=alphabet):
    '''Takes an integer and returns it in the given base with mappings for upper/lower case letters and numbers 0-9.'''
    try:
        i = int(i)
    except Exception:
        raise TypeError("Input must be an integer.")

    def incode(i=i, p=1, a=a):
        # Here to protect p.                                                                                                                                                                                                                
        if i <= 61:
            return lookup(i)

        else:
            pval = pow(62,p)
            nval = i/pval
            remainder = i % pval
            if nval <= 61:
                return lookup(nval) + incode(i % pval)
            else:
                return incode(i, p+1)

    return incode()



def decode(s, a=alphabet):
    '''Takes a base 62 string in our alphabet and returns it in base10.'''
    try:
        s = str(s)
    except Exception:
        raise TypeError("Input must be a string.")

    return sum([lookup(i) * pow(62,p) for p,i in enumerate(list(reversed(s)))])a

这是我的版本，给任何需要的人。

如果你不想重新发明轮子……http://lilurl.sourceforge.net/