How to change the default HTTP OPTIONS parameters in Java

Question

My java snippet looks like:

...
String type = "text/plain;charset=UTF-8";
URL url = new URL("http://xxx/");
HttpURLConnection conn = (HttpURLConnection) url.openConnection();

conn.setDoOutput(true);
conn.setRequestMethod("OPTIONS");
conn.setRequestProperty("Content-Type", type);
...

When I sniff what this sends it sends a

OPTIONS / HTTP/1.1

which appears to be the default.

However, I actually want to send

OPTIONS * HTTP/1.0

How would I do this?

BalusC · Accepted Answer · 2010-02-03 22:51:27Z

3

You can't do that with "plain" java.net.URLConnection. Consider replacing by Apache Commons HttpClient which is less bloated and more configureable. You can force HTTP 1.0 mode by setting http.protocol.version to HttpVersion.HTTP_1_0 in HttpClient#getParams(). You can find an example in this document.

answered Feb 3, 2010 at 22:51

BalusC

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5 Comments

rbrayb Over a year ago

OK - thanks - does the Apache client support the "*" parameter and how would I add it?

BalusC Over a year ago

It's actually the request URI. I haven't used this particular method, so don't pinpoint me on it, but there's a OptionsMethod class taking an URI. You could pass * to it: hc.apache.org/httpclient-3.x/apidocs/org/apache/commons/…

Buhake Sindi Over a year ago

On HttpClient 4.0 there's no OptionsMethod, it's rather HttpOptions class (hc.apache.org/httpcomponents-client/httpclient/apidocs/org/…)

BalusC Over a year ago

@The Elite Gentleman: That's another (newer) API :) Click at the link I provided in my comment. A good hint though, upgrading to HttpClient 4.0.

Buhake Sindi Over a year ago

I know, on Apache sites, they've mentioned that we should rather use HttpClient 4.0 or higher since they've made it more flexible that it's previous releases. :-)

Dilip Rajkumar · Accepted Answer · 2012-04-28 13:17:12Z

2

I agree with the answer the following is the code using HTTPClient

HttpClient client = new DefaultHttpClient(); 
            client.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_0);

Hope it helps some one..

answered Apr 28, 2012 at 13:17

Dilip Rajkumar

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