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寫這種程序真是動力十足啊，哈哈  回復  更多評論
  

用技術創造快感,呵呵  回復  更多評論
  

這個就是動力啊！  回復  更多評論
  

問個問題，當有代理的時候，我用HttpClient 后，用 HttpHost targetHost = new HttpHost(url);
HttpHost proxy = new HttpHost("10.231.96.132", 808);
httpclient.getParams().setParameter(ConnRoutePNames.DEFAULT_PROXY, proxy);

HttpGet httpget = new HttpGet("");
HttpResponse response = httpclient.execute(targetHost, httpget);
執行后 System.out.println(response.getStatusLine()); 出現HTTP/1.1 404 /bbs.wed114.cn/data/attachment/forum/201105/08/174414bfl5qf6qfln3q7g3.jpg/& 后面老多個/& 怎么處理。
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2.7. HttpClient proxy configuration
Even though HttpClient is aware of complex routing scemes and proxy chaining, it supports only simple direct or one hop proxy connections out of the box.

The simplest way to tell HttpClient to connect to the target host via a proxy is by setting the default proxy parameter:

DefaultHttpClient httpclient = new DefaultHttpClient();

HttpHost proxy = new HttpHost("someproxy", 8080);
httpclient.getParams().setParameter(ConnRoutePNames.DEFAULT_PROXY, proxy);

One can also instruct HttpClient to use the standard JRE proxy selector to obtain proxy information:

DefaultHttpClient httpclient = new DefaultHttpClient();

ProxySelectorRoutePlanner routePlanner = new ProxySelectorRoutePlanner(
httpclient.getConnectionManager().getSchemeRegistry(),
ProxySelector.getDefault());
httpclient.setRoutePlanner(routePlanner);

Alternatively, one can provide a custom RoutePlanner implementation in order to have a complete control over the process of HTTP route computation:

DefaultHttpClient httpclient = new DefaultHttpClient();
httpclient.setRoutePlanner(new HttpRoutePlanner() {

public HttpRoute determineRoute(
HttpHost target,
HttpRequest request,
HttpContext context) throws HttpException {
return new HttpRoute(target, null, new HttpHost("someproxy", 8080),
"https".equalsIgnoreCase(target.getSchemeName()));
}

});

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