我想创建一个URL缩短服务,你可以写一个长URL到输入字段和服务缩短URL为“http://www.example.org/abcdef”。

可以用包含a-z, a-z和0-9的6个字符的字符串代替"abcdef"。这样就有560 ~ 570亿个可能的字符串。

我的方法:

我有一个有三列的数据库表:

Id,整数,自动递增 long, string,用户输入的长URL 短,字符串,缩短的URL(或只有六个字符)

然后将长URL插入到表中。然后,我将为“id”选择自动递增的值,并构建它的散列。这个散列应该作为“short”插入。但是我应该构建什么样的哈希呢?像MD5这样的哈希算法会创建太长的字符串。我认为我不用这些算法。一个自建的算法也可以。

我的想法:

对于“http://www.google.de/”,我得到了自动增量id 239472。然后我执行以下步骤:

short = '';
if divisible by 2, add "a"+the result to short
if divisible by 3, add "b"+the result to short
... until I have divisors for a-z and A-Z.

可以不断重复,直到这个数不能被整除为止。你认为这是一个好方法吗?你有更好的主意吗?

由于对这个主题的持续兴趣,我已经发布了一个高效的GitHub解决方案,包括JavaScript、PHP、Python和Java的实现。如果你喜欢,添加你的解决方案:)


当前回答

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。

其他回答

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)
  }

}
// simple approach

$original_id = 56789;

$shortened_id = base_convert($original_id, 10, 36);

$un_shortened_id = base_convert($shortened_id, 36, 10);

c#版本:

public class UrlShortener 
{
    private static String ALPHABET = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
    private static int    BASE     = 62;

    public static String encode(int num)
    {
        StringBuilder sb = new StringBuilder();

        while ( num > 0 )
        {
            sb.Append( ALPHABET[( num % BASE )] );
            num /= BASE;
        }

        StringBuilder builder = new StringBuilder();
        for (int i = sb.Length - 1; i >= 0; i--)
        {
            builder.Append(sb[i]);
        }
        return builder.ToString(); 
    }

    public static int decode(String str)
    {
        int num = 0;

        for ( int i = 0, len = str.Length; i < len; i++ )
        {
            num = num * BASE + ALPHABET.IndexOf( str[(i)] ); 
        }

        return num;
    }   
}

不知道是否有人会发现这个有用-它更像是一个“hack n斜杠”的方法,但如果你只想要特定的字符,它很简单,工作得很好。

$dictionary = "abcdfghjklmnpqrstvwxyz23456789";
$dictionary = str_split($dictionary);

// Encode
$str_id = '';
$base = count($dictionary);

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


// Decode
$id_ar = str_split($str_id);
$id = 0;

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

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;
}