0

这是我的php源代码：

<?php
    $path = '/images/one.jpg';
    $imm = 'http://www.allamoda.eu/wp-content/uploads/2012/05/calzedonia_290x435.jpg';
    if( $content = file_get_contents($imm) ){   
    file_put_contents($path, $content);   
    echo "Yes";
    }else{
    echo "No";
    }
?>

我得到这个错误：

Warning: file_get_contents(http://www.allamoda.eu/wp-content/uploads/2012/05/calzedonia_290x435.jpg) [function.file-get-contents]: failed to open stream: HTTP request failed! HTTP/1.1 403 Forbidden in /opt/lampp/htdocs/test/down.php on line 4
No

为什么 ？

4

4 回答 4

1

服务器需要一些标头（尤其是Accept和User-Agent）。使用stream_context- 参数file_get_contents()来提供它们：

<?php
$path = '/images/one.jpg';
$opts = array(
  'http'=>array(
    'method'=>"GET",
    'header'=>"Accept-language: en\r\n" .
              "Accept:image/png,image/*;q=0.8,*/*;q=0.5 \r\n".
              "Host: www.allamoda.eu\r\n" .
              "User-Agent: Mozilla/5.0 (Windows NT 5.1; rv:12.0) Gecko/20100101 Firefox/12.0\r\n"
  )
);

    $context = stream_context_create($opts);
    $imm = 'http://www.allamoda.eu/wp-content/uploads/2012/05/calzedonia_290x435.jpg';
    if( $content = file_get_contents($imm,false,$context) ){   

    file_put_contents($path, $content);   
    echo "Yes";

    }else{
    echo "No";
    }
?>
于 2012-05-16T11:23:21.100 回答
0

你不能下载这个文件，服务器allamoda.eu说 (HTTP 403)。

于 2012-05-16T11:05:29.993 回答
0

代码没有错。服务器根本不让你（要么你有太多的请求，要么它只是阻止所有脚本抓取它）。

于 2012-05-16T11:06:53.740 回答
0

您不能直接打开文件。但是您可以尝试使用套接字来获取它的内容：

function getRemoteFile($url)
{
   // get the host name and url path
   $parsedUrl = parse_url($url);
   $host = $parsedUrl['host'];
   if (isset($parsedUrl['path'])) {
      $path = $parsedUrl['path'];
   } else {
      // the url is pointing to the host like http://www.mysite.com
      $path = '/';
   }

   if (isset($parsedUrl['query'])) {
      $path .= '?' . $parsedUrl['query'];
   } 

   if (isset($parsedUrl['port'])) {
      $port = $parsedUrl['port'];
   } else {
      // most sites use port 80
      $port = '80';
   }

   $timeout = 10;
   $response = '';
   // connect to the remote server 
   $fp = @fsockopen($host, '80', $errno, $errstr, $timeout );

   if( !$fp ) { 
      echo "Cannot retrieve $url";
   } else {
      // send the necessary headers to get the file 
      fputs($fp, "GET $path HTTP/1.0\r\n" .
                 "Host: $host\r\n" .
                 "User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.0.3) Gecko/20060426 Firefox/1.5.0.3\r\n" .
                 "Accept: */*\r\n" .
                 "Accept-Language: en-us,en;q=0.5\r\n" .
                 "Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7\r\n" .
                 "Keep-Alive: 300\r\n" .
                 "Connection: keep-alive\r\n" .
                 "Referer: http://$host\r\n\r\n");

      // retrieve the response from the remote server 
      while ( $line = fread( $fp, 4096 ) ) { 
         $response .= $line;
      }

      fclose( $fp );

      // strip the headers
      $pos      = strpos($response, "\r\n\r\n");
      $response = substr($response, $pos + 4);
   }

   // return the file content 
   return $response;
}

例子：

$content = getRemoteFile('http://www.allamoda.eu/wp-content/uploads/2012/05/calzedonia_290x435.jpg');

资源

于 2012-05-16T11:09:59.547 回答