用PHP实现ROT13

我在阅读了有关Jon Skeet的有趣内容之后找到了一个字符串,我猜这是在ROT13中.在检查我的猜测之前,我想我会尝试用PHP解密它.这就是我的所作所为：

$string = "Vs lbh nfxrq Oehpr Fpuarvre gb qrpelcg guvf, ur'q pehfu lbhe fxhyy jvgu uvf ynhtu.";
$tokens = str_split($string);
for ($i = 1; $i <= sizeof($tokens); $i++) {
    $char = $tokens[$i-1];
    for ($c = 1; $c <= 13; $c++) {
        $char++;
    }
    echo $char;
}

我的字符串回来了,如你自己aabakaead ABruacae Sacahnaeaiaer到adaeacrypt tahais,ahae’ad acrusah你的sakualal waitah ahais alaauagah.

我的逻辑似乎很接近,但显然是错的.你能帮帮我吗？

解决方法:

这是一个有效的实现,不使用嵌套循环.您也不需要将字符串拆分为数组,因为您可以像使用PHP中的字符串数组一样索引单个字符.

您需要知道ASCII大写字符的范围是65 – 99,小写字符的范围是97 – 122.如果当前字符在其中一个范围内,则将ASCII值加13.然后,检查是否应该转到字母表的开头.如果你应该翻身,减去26.

$string = "Vs lbh nfxrq Oehpr Fpuarvre gb qrpelcg guvf, ur'q pehfu lbhe fxhyy jvgu uvf ynhtu.";

for ($i = 0, $j = strlen( $string); $i < $j; $i++) 
{
    // Get the ASCII character for the current character
    $char = ord( $string[$i]); 

    // If that character is in the range A-Z or a-z, add 13 to its ASCII value
    if( ($char >= 65  && $char <= 90) || ($char >= 97 && $char <= 122)) 
    {
        $char += 13; 

        // If we should have wrapped around the alphabet, subtract 26
        if( $char > 122 || ( $char > 90 && ord( $string[$i]) <= 90)) 
        {
            $char -= 26;
        }
    }
    echo chr( $char);
}

这会产生：

If you asked Bruce Schneier to decrypt this, he’d crush your skull with his laugh.

标签：php,encryption,rot13
来源： https://codeday.me/bug/20190714/1455676.html