如何根据RFC2231对PHP中的文件名进行编码？

Question

如何根据MIME Parameter Value and Encoded Word Extensions: Character Sets, Languages, and Continuations (RFC 2231)的编码对文件名的值进行编码

Answer 1

我认为这应该可以做到：

function rfc2231_encode($name, $value, $charset='', $lang='', $ll=78) {
    if (strlen($name) === 0 || preg_match('/[\x00-\x20*\'%()<>@,;:\\\\"\/[\]?=\x80-\xFF]/', $name)) {
        // invalid parameter name;
        return false;
    }
    if (strlen($charset) !== 0 && !preg_match('/^[A-Za-z]{1,8}(?:-[A-Za-z]{1,8})*$/', $charset)) {
        // invalid charset;
        return false;
    }
    if (strlen($lang) !== 0 && !preg_match('/^[A-Za-z]{1,8}(?:-[A-Za-z]{1,8})*$/', $lang)) {
        // invalid language;
        return false;
    }
    $value = "$charset'$lang'".preg_replace_callback('/[\x00-\x20*\'%()<>@,;:\\\\"\/[\]?=\x80-\xFF]/', function($match) { return rawurlencode($match[0]); }, $value);
    $nlen = strlen($name);
    $vlen = strlen($value);
    if (strlen($name) + $vlen > $ll-3) {
        $sections = array();
        $section = 0;
        for ($i=0, $j=0; $i<$vlen; $i+=$j) {
            $j = $ll - $nlen - strlen($section) - 4;
            $sections[$section++] = substr($value, $i, $j);
        }
        for ($i=0, $n=$section; $i<$n; $i++) {
            $sections[$i] = " $name*$i*=".$sections[$i];
        }
        return implode(";\r\n", $sections);
    } else {
        return " $name*=$value";
    }
}

请注意，此函数期望在单独的行中使用输出，前面有适当的换行(即CRLF)，例如：

"Content-Type: application/x-stuff;\r\n".rfc2231_encode('title', 'This is even more ***fun*** isn\'t it!', 'us-ascii', 'en', 48)

输出为：

Content-Type: application/x-stuff;
 title*0*=us-ascii'en'This%20is%20even%20more%20;
 title*1=%2A%2A%2Afun%2A%2A%2A%20isn%27t%20it!

另请参见。