Question

当我试图通过在php中使用file_get_contents从外部url fanpop.com获取网站内容时，我得到空数据。我使用下面的代码来获取内容

$add_url= "http://www.fanpop.com/";  
$add_domain = file_get_contents($add_url);  
echo $add_domain;

但是在这里我得到$ add_domain的空结果。但是相同的代码正在为其他网址工作，我试图从浏览器发送请求而不是从脚本发送，然后它也无法正常工作。

以下是相同的请求，但在CURL中：

error_reporting(-1);
ini_set('display_errors','On');
$url="http://www.fanpop.com/";
$ch = curl_init();

$header=array('GET /1575051 HTTP/1.1',
'Host: adfoc.us',
'Accept:text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
'Accept-Language:en-US,en;q=0.8',
'Cache-Control:max-age=0',
'Connection:keep-alive',
'Host:adfoc.us',
'User-Agent:Mozilla/5.0 (Macintosh; Intel Mac OS X 10_8_4)            AppleWebKit/537.36 (KHTML, like Gecko) Chrome/27.0.1453.116 Safari/537.36',
);

curl_setopt($ch,CURLOPT_URL,$url);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,0);
curl_setopt( $ch, CURLOPT_COOKIESESSION, true );

curl_setopt($ch,CURLOPT_COOKIEFILE,'cookies.txt');
curl_setopt($ch,CURLOPT_COOKIEJAR,'cookies.txt');
curl_setopt($ch,CURLOPT_HTTPHEADER,$header);

echo $ result = curl_exec（$ ch）;

curl_close($ch);

...但是上面也没有用，任何一个人都可以告诉我有什么变化吗？

Answer 1

这个特定网站的问题是它只提供压缩内容，否则会抛出404错误。

轻松修复：

$ch = curl_init('http://www.fanpop.com');
curl_setopt($ch,CURLOPT_ENCODING , "");
curl_exec($ch);

您也可以file_get_contents()为public class LocalHttpServer extends NanoHTTPD { public static final int SERVER_PORT = 5987; private String mUrl; private InputStream input; private FileOutputStream output; public LocalHttpServer(String url) { super(SERVER_PORT); mUrl = url; } private File createFile(String url) { File path = new File(MyApplication.getContext().getFilesDir(), "audio/"); path.mkdirs(); return new File(path, Util.md5(url)); } @Override public Response serve(IHTTPSession session) { input = null; output = null; HttpURLConnection connection = null; try { URL url = new URL(mUrl); connection = (HttpURLConnection) url.openConnection(); connection.connect(); if (connection.getResponseCode() != HttpURLConnection.HTTP_OK) { return new Response(Response.Status.BAD_REQUEST, "audio/mpeg3", null, 0); } int fileLength = connection.getContentLength(); input = connection.getInputStream(); output = new FileOutputStream(createFile(mUrl)); new Thread(new Runnable() { @Override public void run() { byte data[] = new byte[4096]; int count; try { while ((count = input.read(data)) != -1) { output.write(data, 0, count); } } catch (IOException e) { e.printStackTrace(); } finally { try { if (output != null) output.close(); if (input != null) //input.close(); don't close it } catch (IOException e) { e.printStackTrace(); } } } }).start(); return new Response(Response.Status.OK, "audio/mpeg3", input, fileLength); } catch (IOException e) { e.printStackTrace(); } return new Response(Response.Status.BAD_REQUEST, "audio/mpeg3", null, 0); } }工作，但需要花费大量精力，如this article中所述。

无法通过在php中使用file_get_content获取网站内容

1 个答案: