Is it possible to get a HttpConnection from HttpClient?

[Yan aka. Yansheng]

Hi There,

This is my first post here so I hope this is the right audience.

My story: I want to replace a chunk of my code for doing I/O streaming to a
webservice.  I use java.net.HttpURLConnection to get the InputStream and
OutputStream, it works fine most of the time but it can get a bit flaky and
I had to write my own retry logic.  So I decided to switch to
jakartacommons-httpclient to see if things get better.  The problem is I
don't seem to be able to come up with a good replacement from the httpclient
package.

Here is the method for obtaining a HttpConnection I wrote, the problem is
the HostConfiguration returned is not populated.

--------
    private HttpConnection openConnection(URL url) throws IOException {
        HttpMethod method = new PostMethod(url.toString());
        HttpConnection conn = null;

        // Execute the method.
        int statusCode = getHttpClient().executeMethod(method);

        if (statusCode != HttpStatus.SC_OK) {
            method.releaseConnection();
            throw new IOException("Unable to connect to Service: " + url);
        }
        else {
            HttpConnectionManager connManager =
client.getHttpConnectionManager();
            conn = connManager.getConnection(client.getHostConfiguration());
        }

        return conn;
    }
-------

The exception I'm getting:

Exception in thread "main" java.lang.IllegalArgumentException: host
parameter is null
    at
org.apache.commons.httpclient.HttpConnection.setHost(HttpConnection.java:248)
    at
org.apache.commons.httpclient.SimpleHttpConnectionManager.getConnectionWithTimeout(SimpleHttpConnectionManager.java:163)
    at
org.apache.commons.httpclient.SimpleHttpConnectionManager.getConnection(SimpleHttpConnectionManager.java:191)
    at
org.apache.commons.httpclient.SimpleHttpConnectionManager.getConnection(SimpleHttpConnectionManager.java:108)
    at
amazon.wma.storage.timber.TimberServiceImpl.openConnection(MyServiceImpl.java:180)
    ...

Thanks for any help!
Yansheng