3

我想在不依赖 cURL 的情况下发出 HTTP 请求，并allow_url_fopen = 1通过打开套接字连接并发送原始 HTTP 请求：

/**
 * Make HTTP GET request
 *
 * @param   string   the URL
 * @param   int      will be filled with HTTP response status code
 * @param   string   will be filled with HTTP response header
 * @return  string   HTTP response body
 */
function http_get_request($url, &$http_code = '', &$res_head = '') 
{
  $scheme = $host = $user = $pass = $query = $fragment = '';
  $path = '/';
  $port = substr($url, 0, 5) == 'https' ? 443 : 80;

  extract(parse_url($url)); 

  $path .= ($query ? "?$query" : '').($fragment ? "#$fragment" : '');

  $head = "GET $path HTTP/1.1\r\n"
        . "Host: $host\r\n"
        . "Authorization: Basic ".base64_encode("$user:$pass")."\r\n"
        . "Connection: close\r\n\r\n";

  $fp = fsockopen($scheme == 'https' ? "ssl://$host" : $host, $port) or 
    die('Cannot connect!');

  fputs($fp, $head);
  while(!feof($fp)) {
    $res .= fgets($fp, 4096);
  }
  fclose($fp);

  list($res_head, $res_body) = explode("\r\n\r\n", $res, 2);
  list(, $http_code, ) = explode(' ', $res_head, 3);

  return $res_body;
}

该函数工作正常，但由于我使用的是 HTTP/1.1，因此响应正文通常以块编码字符串返回。例如（来自维基百科）：

25
This is the data in the first chunk

1C
and this is the second one

3
con
8
sequence
0

我不想使用http_chunked_decode()，因为它具有 PE​​CL 依赖项，并且我想要一个高度可移植的代码。

如何轻松解码 HTTP 分块编码字符串，以便我的函数可以返回原始 HTML？我还必须确保解码字符串的长度与Content-Length:标题匹配。

任何帮助，将不胜感激。谢谢。

4

3 回答 3

12

由于该函数返回 HTTP 响应标头，因此您应该检查是否'Transfer-Encoding'是'chunked'然后解码分块编码的字符串。在伪代码中：

CALL parse_http_header
IF 'Transfer-Encoding' IS 'chunked'
  CALL decode_chunked

解析 HTTP 响应头：

下面是将 HTTP 响应标头解析为关联数组的函数。

function parse_http_header($str) 
{
  $lines = explode("\r\n", $str);
  $head  = array(array_shift($lines));
  foreach ($lines as $line) {
    list($key, $val) = explode(':', $line, 2);
    if ($key == 'Set-Cookie') {
      $head['Set-Cookie'][] = trim($val);
    } else {
      $head[$key] = trim($val);
    }
  }
  return $head;
}

该函数将返回一个像这样的数组：

Array
(
    [0] => HTTP/1.1 200 OK
    [Expires] => Tue, 31 Mar 1981 05:00:00 GMT
    [Content-Type] => text/html; charset=utf-8
    [Transfer-Encoding] => chunked
    [Set-Cookie] => Array
        (
            [0] => k=10.34; path=/; expires=Sat, 09-Jun-12 01:58:23 GMT; domain=.example.com
            [1] => guest_id=v1%3A13; domain=.example.com; path=/; expires=Mon, 02-Jun-2014 13:58:23 GMT
        )
    [Content-Length] => 43560
)

注意Set-Cookie标题如何解析为数组。您需要稍后解析 cookie 以将 URL 与需要发送的 cookie 相关联。

解码分块编码的字符串

下面的函数将分块编码的字符串作为参数，并返回解码后的字符串。

function decode_chunked($str) {
  for ($res = ''; !empty($str); $str = trim($str)) {
    $pos = strpos($str, "\r\n");
    $len = hexdec(substr($str, 0, $pos));
    $res.= substr($str, $pos + 2, $len);
    $str = substr($str, $pos + 2 + $len);
  }
  return $res;
}

// Given the string in the question, the function above will returns:
//
// This is the data in the first chunk
// and this is the second one
// consequence
于 2012-06-02T02:45:06.410 回答
2

我不知道它是否最适合您需要做的事情，但是如果您指定HTTP/1.0而不是HTTP/1.1，您将不会得到分块响应。

于 2012-05-29T04:24:34.743 回答
1

这个函数在 Wordpress 中使用。

function decode_chunked($data) {
    if (!preg_match('/^([0-9a-f]+)(?:;(?:[\w-]*)(?:=(?:(?:[\w-]*)*|"(?:[^\r\n])*"))?)*\r\n/i', trim($data))) {
        return $data;
    }



    $decoded = '';
    $encoded = $data;

    while (true) {
        $is_chunked = (bool) preg_match('/^([0-9a-f]+)(?:;(?:[\w-]*)(?:=(?:(?:[\w-]*)*|"(?:[^\r\n])*"))?)*\r\n/i', $encoded, $matches);
        if (!$is_chunked) {
            // Looks like it's not chunked after all
            return $data;
        }

        $length = hexdec(trim($matches[1]));
        if ($length === 0) {
            // Ignore trailer headers
            return $decoded;
        }

        $chunk_length = strlen($matches[0]);
        $decoded .= substr($encoded, $chunk_length, $length);
        $encoded = substr($encoded, $chunk_length + $length + 2);

        if (trim($encoded) === '0' || empty($encoded)) {
            return $decoded;
        }
    }

    // We'll never actually get down here
    // @codeCoverageIgnoreStart
}
于 2017-04-07T20:31:30.233 回答