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

我的方法是:获取数据库ID，然后对其进行Base36编码。我不会同时使用大写字母和小写字母，因为这会使通过电话传输这些url成为一场噩梦，但您当然可以轻松地将该功能扩展为62 en/解码器。

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));
    }
}

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")

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

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