为什么不直接生成一个随机字符串并将其附加到基本URL?这是一个非常简化的c#版本。

static string chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
static string baseUrl = "https://google.com/";

private static string RandomString(int length)
{
    char[] s = new char[length];
    Random rnd = new Random();
    for (int x = 0; x < length; x++)
    {
        s[x] = chars[rnd.Next(chars.Length)];
    }
    Thread.Sleep(10);

    return new String(s);
}

然后添加一个随机字符串到baseURL:

string tinyURL = baseUrl + RandomString(5);

记住，这是一个非常简化的版本，RandomString方法可能会创建重复的字符串。在生产中，您需要考虑重复的字符串，以确保始终拥有唯一的URL。我有一些代码，通过查询我可以共享的数据库表来考虑重复的字符串，如果任何人感兴趣的话。

public class TinyUrl {
    
        private final String characterMap = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
        private final int charBase = characterMap.length();
    
        public String covertToCharacter(int num){
            StringBuilder sb = new StringBuilder();
    
            while (num > 0){
                sb.append(characterMap.charAt(num % charBase));
                num /= charBase;
            }
    
            return sb.reverse().toString();
        }
    
        public int covertToInteger(String str){
            int num = 0;
            for(int i = 0 ; i< str.length(); i++)
                num += characterMap.indexOf(str.charAt(i)) * Math.pow(charBase , (str.length() - (i + 1)));
    
            return num;
        }
}
    
class TinyUrlTest{
    
    public static void main(String[] args) {
        TinyUrl tinyUrl = new TinyUrl();
        int num = 122312215;
        String url = tinyUrl.covertToCharacter(num);
        System.out.println("Tiny url:  " + url);
        System.out.println("Id: " + tinyUrl.covertToInteger(url));
    }
}

你是故意省略O 0和i的吗?

我刚刚基于Ryan的解决方案创建了一个PHP类。

<?php

    $shorty = new App_Shorty();

    echo 'ID: ' . 1000;
    echo '<br/> Short link: ' . $shorty->encode(1000);
    echo '<br/> Decoded Short Link: ' . $shorty->decode($shorty->encode(1000));


    /**
     * A nice shorting class based on Ryan Charmley's suggestion see the link on Stack Overflow below.
     * @author Svetoslav Marinov (Slavi) | http://WebWeb.ca
     * @see http://stackoverflow.com/questions/742013/how-to-code-a-url-shortener/10386945#10386945
     */
    class App_Shorty {
        /**
         * Explicitly omitted: i, o, 1, 0 because they are confusing. Also use only lowercase ... as
         * dictating this over the phone might be tough.
         * @var string
         */
        private $dictionary = "abcdfghjklmnpqrstvwxyz23456789";
        private $dictionary_array = array();

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

        /**
         * Gets ID and converts it into a string.
         * @param int $id
         */
        public function encode($id) {
            $str_id = '';
            $base = count($this->dictionary_array);

            while ($id > 0) {
                $rem = $id % $base;
                $id = ($id - $rem) / $base;
                $str_id .= $this->dictionary_array[$rem];
            }

            return $str_id;
        }

        /**
         * Converts /abc into an integer ID
         * @param string
         * @return int $id
         */
        public function decode($str_id) {
            $id = 0;
            $id_ar = str_split($str_id);
            $base = count($this->dictionary_array);

            for ($i = count($id_ar); $i > 0; $i--) {
                $id += array_search($id_ar[$i - 1], $this->dictionary_array) * pow($base, $i - 1);
            }
            return $id;
        }
    }
?>

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。

对于一个类似的项目，为了获得一个新键，我围绕一个随机字符串生成器创建了一个包装器函数，该函数调用生成器，直到我得到一个尚未在哈希表中使用过的字符串。一旦名称空间开始满了，这个方法就会变慢，但正如您所说的，即使只有6个字符，也有足够的名称空间可以使用。

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。