自定义加密算法的实现

由于要传一个需要保密的ID，因此用到对称加密，但mcrypt_encrypt算法加密后字符串太长，因此想实现一个自定义加密算法，想法如下

首先先对key计算sha1，取结果的前32bit，然后跟要加密整数进行异或，得到一个加密后的32bit结果

对结果分组：2bit | 6bit | 6bit | 6bit | 6bit | 6bit

各个组分别取名为：a0、a1、a2、a3、a4、a5

另定义一个长度64的字典数组

$dict=array('1','2','3','4','5','6','7','8','9',
'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',
'01','02','03');

将前面每个分组的值作为字典数组的下标，则加密结果为：$dict[a0].$dict[a1].$dict[a2].$dict[a3].$dict[a4].$dict[a5]

这样加密后的结果就是一个长度6-12的字符串，如果字典数组最后3个元素用其他单字符表示，那么结果就固定为6个字符的字符串。

由于初学php不久，对php的函数库不熟悉，求大侠帮忙实现下加密解密算法：

string encrypt(int id,string key)

int decrypt(string text,string key)

回复讨论(解决方案)

echo encrypt(1234, 'abc'), PHP_EOL;echo decrypt( '1TgGSY', 'abc');function encrypt($id, $key) {  $dict = array('1','2','3','4','5','6','7','8','9',    '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',    '-','=','*'  );  $key = current(unpack('L', sha1($key, 1)));  $id ^= $key;  $t = str_split(sprintf('%036b', $id), 6);  foreach($t as &$v) $v = $dict[bindec($v)];  return join($t);}function decrypt($s, $key) {  $dict = array('1','2','3','4','5','6','7','8','9',    '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',    '-','=','*'  );  $dict = array_flip($dict);  foreach(str_split($s) as $c) $r[] = sprintf('%06b', $dict[$c]);  $id = bindec(join($r));  $key = current(unpack('L', sha1($key, 1)));  return $id ^ $key;}

1TgGSY
1234

我自己也实现了加密过程，不过看起来就没那么优雅了，执行效率也低点，贴出来衬托下高手风范

$key_string = 'abc';function keyToInt($key) {	$key_sha1 = sha1 ( $key );	$first_char = $key_sha1 [0];	if (ord ( $first_char ) > 55) {		return hexdec ( (hexdec ( $first_char ) & 7) . substr ( $key_sha1, 1, 7 ) ) | (- 2147483648);	} else {		return hexdec ( $key_sha1 );	}}function Encrypt($num){		$dict = array('0','1','2','3','4','5','6','7','8','9','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','*','!');	$r1 = $num ^ keyToInt ( $key_string );	$r2 = decbin ( $r1 );	$r3 = array (			bindec ( substr ( $r2, 0, 2 ) ),			bindec ( substr ( $r2, 2, 6 ) ),			bindec ( substr ( $r2, 8, 6 ) ),			bindec ( substr ( $r2, 14, 6 ) ),			bindec ( substr ( $r2, 20, 6 ) ),			bindec ( substr ( $r2, 26, 6 ) ) 	);	return $dict [$r3 [0]] . $dict [$r3 [1]] . $dict [$r3 [2]] . $dict [$r3 [3]] . $dict [$r3 [4]] . $dict [$r3 [5]];}

算法实现之后，发觉在设计算法时，有个缺陷没考虑到，

由于仅仅是id与key异或，加密后的结果存在一定规律性，

比如
1234->1TgGSY
1235->1TgGSX

有没有什么好的办法打散下结果?

这个够乱的了吧

$id = 1234;$key = 'aaa';for($i=1; $i&$v) {    $v = $dict[bindec($v)];    if($i == 0) shuffle($dict);  }  return join($t);}function decrypt($s, $key) {  $dict = array('1','2','3','4','5','6','7','8','9',    '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',    '-','=','*'  );  $m = array_search($s{0}, $dict);  $n = $m >> 2;  srand($n);  shuffle($dict);  $dict = array_flip($dict);  foreach(str_split($s) as $i=>$c) {    $r[] = sprintf('%06b', $i==0 ? $m&0x03 : $dict[$c]);  }      $id = bindec(join($r));  $key = current(unpack('L', substr(sha1($key, 1), $n)));  return $id ^ $key;}

1234       4rHK4B 12341235       oD2LN* 12351236       wqkf8u 12361237       6k=GVU 12371238       bxeCr* 12381239       =W-AOi 12391240       IiQ3e1 12401241       z6uMMA 12411242       WLcnd8 12421243       Rizj*M 12431244       4rHK47 12441245       oD2LNT 12451246       wqkf8Z 12461247       6k=GVJ 12471248       bxeCrE 12481249       =W-AOP 12491250       IiQ3et 12501251       z6uMMP 12511252       WLcndU 12521253       Rizj*p 12531254       4rHK4s 12541255       oD2LNs 12551256       wqkf84 12561257       6k=GVn 12571258       bxeCrL 12581259       =W-AOT 12591260       IiQ3ex 12601261       z6uMM1 12611262       WLcndD 12621263       Rizj*s 12631264       4rHK4h 12641265       oD2LNq 12651266       wqkf83 12661267       6k=GVg 12671268       bxeCr5 12681269       =W-AOH 12691270       IiQ3eP 12701271       z6uMMc 12711272       WLcndE 12721273       Rizj*6 12731274       4rHK4I 12741275       oD2LN= 12751276       wqkf8U 12761277       6k=GVI 12771278       bxeCr9 12781279       =W-AOl 12791280       IiQ3bI 12801281       z6uMhG 12811282       WLcnaY 12821283       Rizj6d 12831284       4rHK3Z 12841285       oD2L*n 12851286       wqkfbP 12861287       6k=Gzj 12871288       bxeC=o 12881289       =W-AEd 12891290       IiQ3bY 12901291       z6uMh* 12911292       WLcnag 12921293       Rizj6v 12931294       4rHK3F 12941295       oD2L*e 12951296       wqkfbJ 12961297       6k=Gzm 12971298       bxeC=N 12981299       =W-AEw 12991300       IiQ3bs 13001301       z6uMhl 13011302       WLcna4 13021303       Rizj6V 13031304       4rHK3u 13041305       oD2L*V 13051306       wqkfbm 13061307       6k=Gz* 13071308       bxeC=- 13081309       =W-AEa 13091310       IiQ3bm 13101311       z6uMhe 13111312       WLcnaS 13121313       Rizj6= 13131314       4rHK38 13141315       oD2L*l 13151316       wqkfbS 13161317       6k=Gz6 13171318       bxeC=q 13181319       =W-AEn 13191320       IiQ3bO 13201321       z6uMhV 13211322       WLcnau 13221323       Rizj61 13231324       4rHK3K 13241325       oD2L*p 13251326       wqkfbv 13261327       6k=Gzw 13271328       bxeC=h 13281329       =W-AE- 13291330       IiQ3bS 13301331       z6uMhj 13311332       WLcna9 1332

在你的设计中，第一节只有 2bit 有效位，所以可在其上再附加4bit信息
而0~15的随机数正好是4bit

算法中，这个随机数起到2个作用
1、调整 key
2、打乱字典

如果不怎么需要太强的保密性，位运算移位就足够了，省点CPU

楼主的解答超赞
加上随机数后，1个整数可对应16种结果，64^6种结果尽数用上，解密时也不用再判断是不是无效字符串了。

是楼主的解答超赞，打错啦

不打了