这个PHP4类做AES加密有什么问题吗？

Question

当我解密使用此函数加密的内容时，解密的版本不等于原始版本。

 class AES256encryption {

    var $secret = '';
    var $cipher_key = '';

    function AES256encryption($secret='') {
        if (empty($secret)) {
            global $secret;         
            if (empty($secret)) {
                $secret = "some random secret string";
            }
        }
        $this->secret = $secret;
    }

    function gen_cipher() {
        if (empty($this->cipher_key)) {
            $this->cipher_key = substr(sha1($this->secret),0,20);
        }
    }
    function mciv() {
        return mcrypt_create_iv(mcrypt_get_iv_size(MCRYPT_RIJNDAEL_256, MCRYPT_MODE_CBC), MCRYPT_RAND);
    }
    function encrypt($text) {
        $this->gen_cipher();
        return trim(base64_encode(mcrypt_encrypt(MCRYPT_RIJNDAEL_256, $this->cipher_key, $text, MCRYPT_MODE_CBC, $this->mciv()))); 
    }
    function decrypt($text) {
        $this->gen_cipher();
        return trim(mcrypt_decrypt(MCRYPT_RIJNDAEL_256, $this->cipher_key, base64_decode($text), MCRYPT_MODE_CBC, $this->mciv())); 
    }
}

Answer 1

不要每次想要加密/解密某些内容时都创建一个新的 IV。加密和解密时需要相同的 IV。在CBC模式下，只要IV秘密在创建时是随机的，就不需要获取IV秘密。所以你的代码应该是这样的：

class AES256encryption {

    var $secret = '';
    var $cipher_key = '';
    var $mciv = NULL;

    function AES256encryption($secret='') {
        if (empty($secret)) {
            global $secret;⋅⋅⋅⋅⋅⋅⋅⋅⋅
            if (empty($secret)) {
                $secret = "some random secret string";
            }
        }
        $this->secret = $secret;
        $this->gen_mciv();
    }   

    function gen_cipher() {
        if (empty($this->cipher_key)) {
            $this->cipher_key = substr(sha1($this->secret),0,20);
        }   
    }   

    function gen_mciv() {
        if(NULL === $this->mciv)
        {
            $this->mciv = mcrypt_create_iv(mcrypt_get_iv_size(MCRYPT_RIJNDAEL_256, MCRYPT_MODE_CBC), MCRYPT_RAND);
        }   
    }   

    function encrypt($text) {
        $this->gen_cipher();
        return trim(base64_encode(mcrypt_encrypt(MCRYPT_RIJNDAEL_256, $this->cipher_key, $text, MCRYPT_MODE_CBC, $this->mciv)));
    }   
    function decrypt($text) {
        $this->gen_cipher();
        return trim(mcrypt_decrypt(MCRYPT_RIJNDAEL_256, $this->cipher_key, base64_decode($text), MCRYPT_MODE_CBC, $this->mciv));
    }   
}   



$ac = new AES256encryption('my secret pass');
$z = $ac->encrypt('test');
var_dump($z);
$u = $ac->decrypt($z);
var_dump($u);

这似乎有效：

mycroft:~ $ php test_aes.php 
string(44) "+KRlfrPp37FfwB4gJXQ67X+8bjbjxEFHjOn55YOgU5o="
string(4) "test"

请检查块密码操作模式 它恢复了它的工作原理。

Answer 2

IV 需要与加密数据一起发送给接收者。这意味着您的加密函数将需要对其进行base64编码并发送它，并且您的解密函数将需要期望接收它作为输入的一部分。

Answer 3

在帕特里克和咖啡馆的帮助下，我修改了我的课程。我发现秘密和 IV 必须与解密中使用的相同加密，否则解密将无法进行。 IV 必须为 32 个字符。这是我修改后的课程，以防对任何人有用。

class AES256 {

    var $secret = 'some string of any length'; // some random string of any length
    var $iv = '0v6bJhPYe2TElCUrT{TD-drLH(5y4pQj'; // must be 32 chars
    var $cipher_key = '';

    function AES256($secret='', $iv='') {
        if (!empty($secret)) {
            $this->secret = $secret;
        }
        $this->cipher_key = substr(sha1($this->secret),0,20);
        if (!empty($iv) && (strlen($iv) == 32)) {
            $this->iv = $iv;
        }
     }
    function encrypt($plaintext) {
         return trim(base64_encode(mcrypt_encrypt(MCRYPT_RIJNDAEL_256, $this->cipher_key, $plaintext, MCRYPT_MODE_CBC, $this->iv))); 
    }
    function decrypt($ciphertext) {
         return trim(mcrypt_decrypt(MCRYPT_RIJNDAEL_256, $this->cipher_key, base64_decode($ciphertext), MCRYPT_MODE_CBC, $this->iv)); 
    }
}

$r = array();

$ac = new AES256('some string of any length');
$r['ciphertext'] = $ac->encrypt(',23ln1gQ6-3ZY[JI');
$r['plaintext'] = $ac->decrypt("wdkUJRR1qxXLkeseVfiLhKnXsAiVzx4H2ytj+2BFRlo=");
print_r($r);