当我们要将一个庞大的数据进行编号时,而编号有位数限制,比如5位的车牌号、10位的某证件号码、订单流水号、短网址等等,我们可以使用36进制计算出符合位数的不重复的编号。
我们将0-Z(0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ)分别代表数值0-35,如字母Z代表35。这样的话我要得到一个5位的编号,最大信息量就是36的5次方了,36^5 = 60466176,即最大的5位编号相当于10进制的数字:60466176。
本文中为了做演示,我们假定某俱乐部发放一批10位的会员卡号,会员卡号由3位城市编号+5位卡号编码+2位校验码组成。城市编号用区号表示,如755代表深圳,5位卡编号则由36进制的卡编号组成,后面两位校验码则是通过一定的算法生成的,校验码的用处是可以验证卡号的合法性。这样的话,我们生成的10位卡号相当于最大能满足6000多万会员卡号,并且是不重复唯一的卡号。
PHP实现
我们使用PHP进行进制转换,10进制转36进制。
class Code {
//密码字典
private $dic = array(
0=>'0', 1=>'1', 2=>'2', 3=>'3', 4=>'4', 5=>'5', 6=>'6', 7=>'7', 8=>'8',
9=>'9', 10=>'A', 11=>'B', 12=>'C', 13=>'D', 14=>'E', 15=>'F', 16=>'G', 17=>'H',
18=>'I',19=>'J', 20=>'K', 21=>'L', 22=>'M', 23=>'N', 24=>'O', 25=>'P', 26=>'Q',
27=>'R',28=>'S', 29=>'T', 30=>'U', 31=>'V', 32=>'W', 33=>'X', 34=>'Y', 35=>'Z'
);
public function encodeID($int, $format=8) {
$dics = $this->dic;
$dnum = 36; //进制数
$arr = array ();
$loop = true;
while ($loop) {
$arr[] = $dics[bcmod($int, $dnum)];
$int = bcdiv($int, $dnum, 0);
if ($int == '0') {
$loop = false;
}
}
if (count($arr) $arr = array_pad($arr, $format, $dics[0]);
return implode('', array_reverse($arr));
}
public function decodeID($ids) {
$dics = $this->dic;
$dnum = 36; //进制数
//键值交换
$dedic = array_flip($dics);
//去零
$id = ltrim($ids, $dics[0]);
//反转
$id = strrev($id);
$v = 0;
for ($i = 0, $j = strlen($id); $i $v = bcadd(bcmul($dedic[$id {
$i }
], bcpow($dnum, $i, 0), 0), $v, 0);
}
return $v;
}
}
我们定义Code类,先定义密码字典,即0-Z分别对应的数值,方法encodeID($int, $format)中参数$int表示数字,$format表示位数长度,比方encodeID(123456789,5)表示将数字123456789转换成5位的36进制编号,而方法decodeID($ids)用于将36进制的编号转换成10进制的编号。
我们可以这样来生成卡号:
$code = new Code();
$card_no = $code->encodeID(888888,5);
如上,我们就可以得到一个5位的卡编号,它实际代表着卡号是888888(6个8)的会员编号,而实际进行转换后是5位编号:0J1VC。
接着,我们将城市编号和校验码加上,城市编号是已经定义好的,校验码则通过一定的算法取得,本例中,我们使用简单的算法:将前三位城市编号和五位卡编号进行md5加密,然后取md5值的前2位作为校验码,这样就得到了编号后面的两位校验码。
$card_pre = '755';
$card_vc = substr(md5($card_pre.$card_no),0,2);
$card_vc = strtoupper($card_vc);
echo $card_pre.$card_no.$card_vc;
实际应用中,可以通过数据库得到10进制的编号,保证编号唯一,再将上述代码组合,最终生成一个10位的不重复的会员卡号。

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