Question

我正在尝试使用Apache HttpClient检索此页面：http://quick-dish.tablespoon.com/

不幸的是，当我尝试这样做时，它只返回以下内容（由JSoup返回，所以可能它只是返回HTTP ...字符串本身）：

<html>
 <head></head>
 <body>
  HTTP/1.1 200 OK [Server: nginx/1.0.11, Content-Type: text/html;charset=UTF-8, Last-Modified: Mon, 02 Jul 2012 15:30:40 GMT, Vary: Accept-Encoding, Cookie,Accept-Encoding, X-Powered-By: PHP/5.3.6, X-Pingback: http://quick-dish.tablespoon.com/xmlrpc.php, X-Powered-By: ASP.NET, Content-Encoding: gzip, X-Blz: lb1.blaze.io, Date: Mon, 02 Jul 2012 16:06:21 GMT, Content-Length: 11723, Connection: keep-alive]
 </body>
</html>

这是我的代码（请注意，我正在仿效Google Bot，因为我发现Web服务器往往表现得更好）：

URL sourceURL = new URL("http://quick-dish.tablespoon.com/");
HttpClient httpClient =  new ContentEncodingHttpClient();
httpClient.getParams().setBooleanParameter("http.protocol.handle-redirects", true);

final HttpGet httpget = new HttpGet(sourceURL.toURI());
httpget.setHeader("User-Agent", "Mozilla/5.0 (compatible; Googlebot/2.1; +http://www.google.com/bot.html)");
httpget.setHeader("Accept", "text/html");
httpget.setHeader("Accept-Charset", "utf-8");

final HttpResponse response = httpClient.execute(httpget);
return Jsoup.parse(response.toString());

毋庸置疑，该页面在我的网络浏览器中返回正常。有什么想法吗？

Answer 1

您需要获取响应实体

，而不是toString

// Get hold of the response entity
 HttpEntity entity = response.getEntity();

然后你可以得到那个

的内容

Answer 2

HttpEntity entity = response.getEntity();
String pageHTML = EntityUtils.toString(entity);
Jsoup.parse(response.toString());

为什么Apache HttpClient 4.2无法检索此页面？

2 个答案: