Retrieving XML over HTTP in Java 7

Question

Some devices (e.g. webrelays) return raw XML in response to HTTPGet requests. That is, the reply contains no valid HTTP header. For many years I have retrieved information from such devices using code like this:

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);
    }
}

In openJdk 7 the following lines have been added to sun.net.www.protocol.http.HttpURLConnection, which mean any HTTP response with an invalid header generates an IOException:

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

How do I get 'headless' XML from a server which expects HTTPGet requests in the new world of Java 7?

Nice catch, though URLConnection is documented to use HTTP1.1 which requires the response to provide a status line in the headers. Only under HTTP1.0 this would be a valid HTTP response. — mabi
– mabi, Commented Nov 13, 2013 at 13:35

Anders R. Bystrup · Accepted Answer · 2013-11-13 13:14:54Z

4

You could always do it the "socket way" and talk HTTP directly to the host:

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():
}

Not exactly elegant, but it'll work. Strange that JRE7 introduces such "regression" though.

Cheers,

edited Nov 13, 2013 at 13:14

answered Nov 13, 2013 at 12:43

Anders R. Bystrup

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1 Comment

Mark Bowman Over a year ago

A beautifully simple solution. Thank you. On my platform (Linux) I had to modify the line termination. So, I used out.print("GET " + relativeURL + " HTTP/1.1 \r\n\r\n");

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