如何根据RFC 2231在PHP中编码文件名？ [英] How can I encode a filename in PHP according to RFC 2231?

问题描述

如何根据 MIME参数值和编码字的编码对文件名的值进行编码扩展：字符集，语言和延续（RFC 2231）？

推荐答案

我认为应该这样做：

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，此函数需要输出用于一个单独的行前面有一个正确的换行符（即CRLF），例如：

Note that this function expects that the output is used in a separate line preceded by a proper line wrap (i.e. CRLF), e.g.:

"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!

参见 HTTP Content-Disposition头字段的测试用例以及RFC 2047和RFC 2231/5987中定义的编码 。

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