如何根據RFC 2231在PHP中對文件名進行編碼？

Question

如何根據MIME Parameter Value and Encoded Word Extensions: Character Sets, Languages, and Continuations (RFC 2231)的編碼對文件名的值進行編碼？

Answer 1

我想這應該這樣做：

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"; 
    } 
} 

注意，這個函數期望的輸出是通過適當的換行（即CRLF）前面有一個單獨的行使用，例如：

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

的輸出是：

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! 

參見Test Cases for HTTP Content-Disposition header field and the Encodings defined in RFC 2047 and RFC 2231/5987。