Question

当我尝试使用base64_decode（）函数

时，我收到以下警告

"Warning: base64_decode() has been disabled for security reasons"

看起来我的主机已禁用base64_ *功能。

我的问题很少

我认为默认情况下可以在php中启用base64_ *函数，对吗？
是否有任何安全原因导致base64_ *功能未启用？任何安全漏洞？
替代base64_ *函数，默认情况下可用吗？
我在哪里可以找到base64_ *实现的自定义类/函数，以便我可以在PHP文件中使用它们，如果PHP的base64_ *函数不可用则使用它们？

帮助表示赞赏。

Answer 1

我为所有base64目的编写了替代函数。

看这里：

function base64_decode($input) {
    $keyStr = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";
    $chr1 = $chr2 = $chr3 = "";
    $enc1 = $enc2 = $enc3 = $enc4 = "";
    $i = 0;
    $output = "";

    // remove all characters that are not A-Z, a-z, 0-9, +, /, or =
    $filter = $input;
    $input = preg_replace("[^A-Za-z0-9\+\/\=]", "", $input);
    if ($filter != $input) {
        return false;
    }

    do {
        $enc1 = strpos($keyStr, substr($input, $i++, 1));
        $enc2 = strpos($keyStr, substr($input, $i++, 1));
        $enc3 = strpos($keyStr, substr($input, $i++, 1));
        $enc4 = strpos($keyStr, substr($input, $i++, 1));
        $chr1 = ($enc1 << 2) | ($enc2 >> 4);
        $chr2 = (($enc2 & 15) << 4) | ($enc3 >> 2);
        $chr3 = (($enc3 & 3) << 6) | $enc4;
        $output = $output . chr((int) $chr1);
        if ($enc3 != 64) {
            $output = $output . chr((int) $chr2);
        }
        if ($enc4 != 64) {
            $output = $output . chr((int) $chr3);
        }
        $chr1 = $chr2 = $chr3 = "";
        $enc1 = $enc2 = $enc3 = $enc4 = "";
    } while ($i < strlen($input));
    return urldecode($output);
}

function base64_encode($data) {
    $b64 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=';
    $o1 = $o2 = $o3 = $h1 = $h2 = $h3 = $h4 = $bits = $i = 0;
    $ac = 0;
    $enc = '';
    $tmp_arr = array();
    if (!$data) {
        return data;
    }
    do {
    // pack three octets into four hexets
    $o1 = charCodeAt($data, $i++);
    $o2 = charCodeAt($data, $i++);
    $o3 = charCodeAt($data, $i++);
    $bits = $o1 << 16 | $o2 << 8 | $o3;
    $h1 = $bits >> 18 & 0x3f;
    $h2 = $bits >> 12 & 0x3f;
    $h3 = $bits >> 6 & 0x3f;
    $h4 = $bits & 0x3f;
    // use hexets to index into b64, and append result to encoded string
    $tmp_arr[$ac++] = charAt($b64, $h1).charAt($b64, $h2).charAt($b64, $h3).charAt($b64, $h4);
    } while ($i < strlen($data));
    $enc = implode($tmp_arr, '');
    $r = (strlen($data) % 3);
    return ($r ? substr($enc, 0, ($r - 3)) . substr('===', $r) : $enc);
}

function charCodeAt($data, $char) {
    return ord(substr($data, $char, 1));
}

function charAt($data, $char) {
    return substr($data, $char, 1);
}

但请勿忘记charCodeAt和charAt功能。它们都是base64_encode所必需的。这两个函数base64_encode和base64_decode都像内置的PHP函数一样工作。但只使用它们，如果内置的那些不可用，因为它们不如内置的那么快。

希望它有所帮助！

Answer 2

刚刚注册以评论jankal的回答，但没有声誉就无法做到。

jankal的答案中的函数base64_encode代码的最后一行是

其中$ r可以是0,1或2.据我所知，代码逻辑的（$ r || 3）表达式值在$ r = 0时必须为3，在其他两种情况下为$ r，但在实践中（PHP 5.6 / 7.0）此表达式的值总是等于1，因此我们的BASE64编码字符串总是以两个＆＃39; =＆＃39;结尾。符号，这绝对是错误的。

我的解决方案：

return ($r ? substr($enc, 0, ($r - 3)) : $enc) . substr('===', ($r || 3));

Answer 3

function myencode($input)
    {
    $CODES = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";

    // the result/encoded string, the padding string, and the pad count
    $r = "";
    $p= "";
    $c = strlen($input) % 3;

    // add a right zero pad to make this string a multiple of 3 characters
    if ($c > 0) 
    {
        for (; $c < 3; $c++)
     {
        $p .= "=";
        $input .= "\0";
     }
    }

    // increment over the length of the string, three characters at a time
    for ($c = 0; $c < strlen($input); $c += 3) {

// we add newlines after every 76 output characters, according to the MIME specs

        if ($c > 0 && ($c / 3 * 4) % 76 == 0)
        $r += "\r\n";

// these three 8-bit (ASCII) characters become one 24-bit number
$n = (ord($input[$c]) << 16) + (ord($input[$c +1]) << 8) + (ord($input[$c +2]));


        // this 24-bit number gets separated into four 6-bit numbers
        $n1 = $n >> 18 & 63;
        $n2 = $n >> 12 & 63;
        $n3 = $n >> 6 & 63;
        $n4 = $n & 63;

 // those four 6-bit numbers are used as indices into the base64 character list

 $r .= "" . $CODES[$n1] . $CODES[$n2] . $CODES[$n3] . $CODES[$n4];

    }
    return substr($r,0,(strlen($r)-strlen($p))) . $p;
    }  



 function mydecode($input) {

 $CODES = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";

// remove/ignore any characters not in the base64 characters list
    // or the pad character -- particularly newlines

    $input= str_replace("[^" . $CODES . "=]","",$input);

    // replace any incoming padding with a zero pad (the 'A' character is  zero)
    $p = ($input[strlen($input) - 1] == '=' ?($input[strlen($input) - 2] == '=' ? "AA" : "A") : "");

    $r = "";
    $input = substr($input,0,(strlen($input) - strlen($p))) . $p;
// increment over the length of this encoded string, four characters at a time
    for ($c = 0; $c < strlen($input); $c += 4) {

// each of these four characters represents a 6-bit index in the
// base64 characters list which, when concatenated, will give the
// 24-bit number for the original 3 characters

         $n=(strpos($CODES,$input[$c]) << 18)
         +  (strpos($CODES,$input[$c+1]) << 12)
         +  (strpos($CODES,$input[$c+2]) << 6)
         +  (strpos($CODES,$input[$c+3]));


// split the 24-bit number into the original three 8-bit (ASCII) characters
$r .= "" . chr(($n >> 16) & 0xFF) . chr((($n >> 8) & 0xFF)) . chr(($n & 0xFF));

    }
    // remove any zero pad that was added to make this a multiple of 24 bits
        return substr($r,0, strlen($r)- strlen($p));
    }

Answer 4

这是托管问题。

请更改您的主机或要求他们删除此安全问题

base64_ *功能已禁用，现在该怎么办？

4 个答案: