在Java 7中通过HTTP检索XML

Question

某些设备(例如webrelay)返回原始XML以响应HTTPGet请求.也就是说,回复不包含有效的HTTP标头.多年来,我使用以下代码从这些设备中检索信息:

private InputStream doRawGET(String url) throws MalformedURLException, IOException
{
    try
    {
        URL url = new URL(url);
        HttpURLConnection con = (HttpURLConnection) url.openConnection();
        con.setConnectTimeout(5000);
        con.setReadTimeout(5000);
        return con.getInputStream();
    }
    catch (SocketTimeoutException ex)
    {
        throw new IOException("Timeout attempting to contact Web Relay at " + url);
    }
}

在openJdk 7中,sun.net.www.protocol.http.HttpURLConnection添加了以下行,这意味着任何带有无效标头的HTTP响应都会生成IOException:

1325                respCode = getResponseCode();
1326                if (respCode == -1) {
1327                    disconnectInternal();
1328                    throw new IOException ("Invalid Http response");
1329                }

如何从Java 7新世界中需要HTTPGet请求的服务器获取"无头"XML？

Answer 1

您始终可以通过“套接字方式”执行此操作，并直接与主机进行 HTTP 通信：

private InputStream doRawGET( String url )
{
    Socket s = new Socket( new URL( url ).getHost() , 80 );
    PrintWriter out = new PrintWriter( s.getOutputStream() , true );
    out.println( "GET " + new URL( url ).getPath() + " HTTP/1.0" );
    out.println(); // extra CR necessary sometimes.
    return s.getInputStream():
}

不太优雅，但它会起作用。奇怪的是，JRE7 引入了这种“回归”。

干杯，