Php base64_*功能已禁用,现在要做什么?
当我尝试使用base64_decode()函数时,收到以下警告 看起来我的主机已禁用base64_*函数 我有几个问题Php base64_*功能已禁用,现在要做什么?,php,base64,decode,encode,Php,Base64,Decode,Encode,当我尝试使用base64_decode()函数时,收到以下警告 看起来我的主机已禁用base64_*函数 我有几个问题 我认为在php中默认情况下可以启用base64_*函数,对吗 没有启用base64_*函数是否有任何安全原因?有安全漏洞吗 默认情况下可用的base64_*函数的替代方案 我在哪里可以找到base64_*实现的自定义类/函数,这样我就可以在我的PHP文件中使用它们,如果PHP的base64_*函数不可用,就可以使用它们 感谢您的帮助。这是托管问题 请更改您的主机或要求他们删除此
感谢您的帮助。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我已经为所有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, $char, 1);
}
函数base64\u解码($input){
$keyStr=“abcdefghijklmnopqrstuvxyzabcdefghijklmnopqrstuvxyz012456789+/=”;
$chr1=$chr2=$chr3=“”;
$enc1=$enc2=$enc3=$enc4=“”;
$i=0;
$output=“”;
//删除所有非A-Z、A-Z、0-9、+、/、或=
$filter=$input;
$input=preg\U替换(“[^A-Za-z0-9\+\/\=]”,“”,$input);
如果($filter!=$input){
返回false;
}
做{
$enc1=strpos($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>4);
$chr2=($enc2&15)>2);
$chr3=($enc3&3)>12&0x3f;
$h3=$bits>>6&0x3f;
$h4=$bits&0x3f;
//使用hexets索引到b64中,并将结果附加到编码字符串中
$tmp_arr[$ac++]=charAt($b64,$h1)、charAt($b64,$h2)、charAt($b64,$h3)、charAt($b64,$h4);
}而($i
但是不要忘记charCodeAt
和charAt
功能。它们都是base64\u encode
所必需的。
这两个函数base64\u encode
和base64\u decode
都像内置PHP函数一样工作。但只有在内置PHP函数由于速度不如内置PHP函数而无法使用时才使用它们
希望有帮助!刚注册评论jankal的答案,但没有声誉就做不到 jankal答案中函数base64_编码代码的最后一行是
return ($r ? substr($enc, 0, ($r - 3)) : $enc) . substr('===', ($r || 3));
其中,$r可以是0、1或2。据我所知,$r=0时,($r | | 3)代码逻辑的表达式值必须是3,在其他两种情况下,$r,但在实践中(PHP5.6/7.0)此表达式的值始终等于1,因此我们的BASE64编码字符串总是以两个“=”符号结尾,这肯定是错误的
我的解决方案:
return ($r ? substr($enc, 0, ($r - 3)) . substr('===', $r) : $enc);
函数myencode($input)
{
$CODES=“abcdefghijklmnopqrstuvxyzabefghijklmnopqrstuvxyzo123456789+/=”;
//结果/编码字符串、填充字符串和填充计数
$r=“”;
$p=“”;
$c=strlen($input)%3;
//添加一个右零填充,使该字符串为3个字符的倍数
如果($c>0)
{
对于(;$c<3;$c++)
{
$p.=“=”;
$input.=“\0”;
}
}
//字符串长度的增量,一次三个字符
对于($c=0;$c0&($c/3*4)%76==0)
$r+=“\r\n”;
//这三个8位(ASCII)字符成为一个24位数字
$n=(作战需求文件($input[$c])18和63;
$n2=$n>>12&63;
$n3=$n>>6&63;
$n4=$n&63;
//这四个6位数字用作base64字符列表的索引
$r.=''.$CODES[$n1]。$CODES[$n2]。$CODES[$n3]。$CODES[$n4];
}
返回substr($r,0,(strlen($r)-strlen($p))。$p;
}
函数mydecode($input){
$CODES=“abcdefghijklmnopqrstuvxyzabefghijklmnopqrstuvxyzo123456789+/=”;
//删除/忽略任何不在base64字符列表中的字符
//或者pad字符,尤其是换行符
$input=str_替换(“[^”。$CODES.=]”,“”,$input);
//将任何传入的填充替换为零填充(“a”字符为零)
$p=($input[strlen($input)-1]='='='?($input[strlen($input)-2]='='='='='?“AA”:“A”):“”);
$r=“”;
$input=substr($input,0,(strlen($input)-strlen($p))。$p;
//此编码字符串长度的增量,一次四个字符
对于($c=0;$c8)和0xFF)).chr($n和0xFF));
}
//移除为使其成为24位的倍数而添加的任何零焊盘
返回substr($r,0,strlen($r)-strlen($p));
}
恶意脚本的分配是基本的64编码到eval函数中。如果你真的需要它,如果你的主人相信“偏执狂的安全”,那就换个主机吧。。。更改主机。他们的间谍软件是否与base64不兼容?:)@哈乔:他们可能认为base64_解码的所有使用都与模糊代码有关。这当然是大错特错了。我使用来自的信息,在PHP()中破解了一个糟糕的base32库。我相信这也不难做到。我现在在工作,但是,如果我下班后感到无聊,我可以尝试添加base64。我爱你!这就是你想在没有时间编码的紧急情况下找到的答案。在紧急情况下,没有时间编码=PThanks@Ivan Shatsky返回($r?substr($enc,0,($r-3))。substr(“==”,$r):$enc)代码>如果需要模拟t
return ($r ? substr($enc, 0, ($r - 3)) . substr('===', $r) : $enc);
function myencode($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));
}