Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
385 views
in Technique[技术] by (71.8m points)

network programming - HttpURLConnection implementation

I have read that HttpURLConnection supports persistent connections, so that a connection can be reused for multiple requests. I tried it and the only way to send a second POST was by calling openConnection for a second time. Otherwise I got a IllegalStateException("Already connected"); I used the following:

try{
URL url = new URL("http://someconection.com");
}
catch(Exception e){}
HttpURLConnection con = (HttpURLConnection) url.openConnection();
//set output, input etc
//send POST
//Receive response
//Read whole response
//close input stream
con.disconnect();//have also tested commenting this out
con = (HttpURLConnection) url.openConnection();
//Send new POST

The second request is send over the same TCP connection (verified it with wireshark) but I can not understand why (although this is what I want) since I have called disconnect. I checked the source code for the HttpURLConnection and the implementation does keep a keepalive cache of connections to the same destinations. My problem is that I can not see how the connection is placed back in the cache after I have send the first request. The disconnect closes the connection and without the disconnect, still I can not see how the connection is placed back in the cache. I saw that the cache has a run method to go through over all idle connections (I am not sure how it is called), but I can not find how the connection is placed back in the cache. The only place that seems to happen is in the finished method of httpClient but this is not called for a POST with a response. Can anyone help me on this?

EDIT My interest is, what is the proper handling of an HttpUrlConnection object for tcp connection reuse. Should input/output stream be closed followed by a url.openConnection(); each time to send the new request (avoiding disconnect())? If yes, I can not see how the connection is being reused when I call url.openConnection() for the second time, since the connection has been removed from the cache for the first request and can not find how it is returned back. Is it possible that the connection is not returned back to the keepalive cache (bug?), but the OS has not released the tcp connection yet and on new connection, the OS returns the buffered connection (not yet released) or something similar? EDIT2 The only related i found was from JDK_KeepAlive

...when the application calls close() on the InputStream returned by URLConnection.getInputStream(), the JDK's HTTP protocol handler will try to clean up the connection and if successful, put the connection into a connection cache for reuse by future HTTP requests.

But I am not sure which handler is this. sun.net.www.protocol.http.Handler does not do any caching as I saw Thanks!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Should input/output stream be closed followed by a url.openConnection(); each time to send the new request (avoiding disconnect())?

Yes.

If yes, I can not see how the connection is being reused when I call url.openConnection() for the second time, since the connection has been removed from the cache for the first request and can not find how it is returned back.

You are confusing the HttpURLConnection with the underlying Socket and its underlying TCP connection. They aren't the same. The HttpURLConnection instances are GC'd, the underlying Socket is pooled, unless you call disconnect().


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...