php - 如何在php中没有mcrypt库的情况下加密字符串

Question

在函数中需要 key 来加密字符串而无需在 php 中使用 mcrypt libraly

function encrypt($str, $pass){
        $str_arr = str_split($str);
        $pass_arr = str_split($pass);
        $add = 0;
        $div = strlen($str) / strlen($pass);
        while ($add <= $div) {
             $newpass .= $pass;
             $add++;
        }
        $pass_arr = str_split($newpass);
        foreach($str_arr as $key =>$asc) {
            $pass_int = ord($pass_arr[$key]);
            $str_int = ord($asc);
            $int_add = $str_int + $pass_int;
            $ascii .= chr($int_add);
        }
        return $ascii;
    }

    function decrypt($enc, $pass){
        $enc_arr = str_split($enc);
        $pass_arr = str_split($pass);
        $add = 0;
        $div = strlen($enc) / strlen($pass);
        while ($add <= $div) {
            $newpass .= $pass;
            $add++;
        }
        $pass_arr = str_split($newpass);
        foreach($enc_arr as $key =>$asc) {
            $pass_int = ord($pass_arr[$key]);
            $enc_int = ord($asc);
            $str_int = $enc_int - $pass_int;
            $ascii .= chr($str_int);
        }
        return $ascii;
    }

这对我测试的角色不起作用

Answer 1

还有一个方法:

$key = 'the quick brown fox jumps over the lazy ';
$string = 'Hey we are testing encryption';

echo enc_encrypt($string, $key)."\n";
echo enc_decrypt(enc_encrypt($string, $key), $key)."\n";

function enc_encrypt($string, $key) {
    $result = '';
    for($i = 0; $i < strlen($string); $i++) {
        $char = substr($string, $i, 1);
        $keychar = substr($key, ($i % strlen($key))-1, 1);
        $char = chr(ord($char) + ord($keychar));
        $result .= $char;
    }

    return base64_encode($result);
}

function enc_decrypt($string, $key) {
    $result = '';
    $string = base64_decode($string);

    for($i = 0; $i < strlen($string); $i++) {
        $char = substr($string, $i, 1);
        $keychar = substr($key, ($i % strlen($key))-1, 1);
        $char = chr(ord($char) - ord($keychar));
        $result .= $char;
    }

    return $result;
}

同样，不要指望这会非常安全。