使用netty4异步调用链接HTTP请求-响应

Question

I am writing a simple HTTP client using netty-4.0x.
Build the pipeline as below :

pipeline.addLast("codec", new HttpClientCodec());
pipeline.addLast("inflater", new HttpContentDecompressor());
pipeline.addLast("handler", new HttpResponseHandler());

where HttpResponseHandler provides implementation of messageReceived() ,

Now there is a thread-pool which call the client and keep sending http message, I understand that ChannelFuture future = channel.write(request); is async call and will come out without blocking

The query which i am having is, is there a way to link request-response, without calling future.sync() call.

Thanks for all the help in advance !!!

Answer 1

如果也对服务器进行编码，则可以让客户端向HTTP标头添加唯一的ID，并让服务器在响应中回显它。

Answer 2

If you are following strict HTTP pipelining rules then, on any given channel, the responses will be returned in the order the requests were sent in. It should be enough to maintain a request queue, removing the front of the queue for each response received.

If you are creating a new channel, and a new pipeline for that channel, for each request, it's even easier. Either way you can add a handler to your pipeline which remembers the request (or queue of requests), and returns the request and response to your application when the response is received.