PHP简短的唯一ID生成使用auto_increment吗? [英] PHP short unique ID generation using auto_increment?

问题描述

我想生成一个简短的唯一ID，而不必检查冲突.

I would like to generate a short, unique ID without having to check for collisions.

我目前正在执行类似的操作，但是我当前生成的ID是随机的，并且在循环中检查冲突很烦人，如果记录数量显着增长，它将变得很昂贵.

I currently do something like this, but the ID I currently generate is random and checking for collisions in a loop is annoying and will get expensive if the number of records grows significantly.

通常不用担心冲突，但是我要生成的唯一ID是一个简短的唯一字符串，由5-8个字母数字组成，就像tinyurl一样.

Normally worrying about collisions isn't an issue, but the unique ID I want to generate is a short unique string 5-8 characters, alpha-numeric, like tinyurl does.

我想以5个字符开头，如果我输入了6000万个条目，则转到6 ..依此类推.

I would like to start out with 5 characters and if I hit 60 million entries, then go to 6.. so on and so forth.

为此，我想我可以使用对用户隐藏的auto_increment值，并用MD5或其他方法向他们显示，以从中生成唯一的字符串.

To this end, I was thinking I could use an auto_increment value that is hidden from the users, and present them instead with an MD5 or some other method to generate a unique string from that.

生成的字符串不应看起来是线性的，因此将auto_incremented ID转换为base 36 [0-9A-Z]有点太简单了，但是类似的功能正是我要解决的问题.

Generated strings should not appear to be linear, so simply converting the auto_incremented ID into base 36 [0-9A-Z] is a bit too simplistic, but a function something like that is where I'm going with this.

安全性不是问题，因为它将不会用于保护信息.它只是较长字符串的快捷方式. 谢谢.

Security is not an issue as this will not be used to secure information. It is simply a shortcut to a longer string. Thank you.

谢谢您的建议，对您的延迟深表歉意.牙医..

Thank you for your suggestions and sorry for the delay. Dentist..

推荐答案

您将需要一些在构造上正确的东西，即置换函数:这是一个对一个整数进行一对一，可逆映射的函数(您的顺序计数器). 一些示例(这些方法的任何组合也都可以使用):

You'll need something that's correct by construction, i.e. a permutation function: this is a function that does a one-to-one, reversible mapping of one integer (your sequential counter) to another. Some examples (any combination of these should also work):

• 反转一些位(f.i.使用XOR，PHP中为^)
• 交换位(($ i& 0xc)>> 2 |($ i& 0x3)<<< 2)，或只是反转所有位的顺序
• 以最大范围为模添加一个常数(如果将其与上述范围相结合，则必须为两个因数)

示例:此函数会将0、1、2、3、5 ...转换为13、4、12、7、15 ..，最大数字为15:

Example: this function will convert 0, 1, 2, 3, 5, .. into 13, 4, 12, 7, 15, .. for numbers up to 15:

$i=($input+97) & 0xf;
$result=((($i&0x1) << 3) + (($i&0xe) >> 1)) ^ 0x5;

编辑

一种更简单的方法是使用线性同余生成器(LCG，通常用于生成随机数)，该生成器由以下形式的公式定义:

An easier way would to use a linear congruential generator (LCG, which is usually used for generating random numbers), which is defined by a formula of the form:

X_n+1 = (a * X_n + c) mod m

对于 a，c和m的好值，它们是X_0，X_1的序列. X_m-1将一次包含0到m-1之间的所有数字.现在，您可以从线性增加的索引开始，并将LCG序列中的 next 值用作秘密"密钥.

For good values of a, c and m, the sequence of X_0, X_1 .. X_m-1 will contain all numbers between 0 and m-1 exactly once. Now you can start from a linearly increasing index, and use the next value in the LCG sequence as your "secret" key.

EDIT2

实施: 您可以设计自己的LCG参数，但是如果弄错了，它将无法涵盖完整范围(因此有重复项)，因此我将在本文:

Implementation: You can design your own LCG parameters, but if you get it wrong it won't cover the full range (and thus have duplicates) so I'll use a published and tried set of parameters here from this paper:

a = 16807, c = 0, m = 2147483647

这给你2 ** 31的范围.使用pack()，您可以将生成的整数作为字符串，base64_encode()使其成为可读的字符串(最多6个有效字符，每个字节6位)，因此这可能是您的功能:

This gives you a range of 2**31. With pack() you can get the resulting integer as a string, base64_encode() makes it a readable string (of up to 6 significant characters, 6 bits per byte) so this could be your function:

substr(base64_encode(pack("l", (16807 * $index) % 2147483647)), 0, 6)

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