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HttpRequest httpReq=new DefaultHttpRequest(HttpVersion.HTTP_1_1,HttpMethod.POST,uri);
httpReq.setHeader(HttpHeaders.Names.HOST,host);
httpReq.setHeader(HttpHeaders.Names.CONNECTION,HttpHeaders.Values.KEEP_ALIVE);
httpReq.setHeader(HttpHeaders.Names.ACCEPT_ENCODING,HttpHeaders.Values.GZIP);
String params="a=b&c=d";
ChannelBuffer cb=ChannelBuffers.copiedBuffer(params,Charset.defaultCharset());
httpReq.setHeader(HttpHeaders.Names.CONTENT_LENGTH,cb.readableBytes());
httpReq.setContent(cb);

不产生有效的请求。发送发布请求的正确方法是什么,最好是手动构造参数数据,而不是使用 DataFactory。另外,为什么 HttpDataFactory 不包含在任何版本中?

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2 回答 2

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你写的一切都是正确的,只需添加httpReq.setHeader(HttpHeaders.Names.CONTENT_TYPE,"application/x-www-form-urlencoded"); ,你的例子就会起作用。对于更复杂的代码,您需要添加 url 编码。

于 2012-02-29T18:07:52.130 回答
0
DefaultFullHttpRequest request = new DefaultFullHttpRequest(HttpVersion.HTTP_1_1, HttpMethod.POST, uri.toASCIIString());
request.headers().set(HttpHeaders.Names.HOST, ip);
request.headers().set(HttpHeaders.Names.CONTENT_TYPE,"application/x-www-form-urlencoded");
List<NameValuePair> nvps = new ArrayList<NameValuePair>();
nvps.add(new BasicNameValuePair(param.getKey(), param.getValue()));
HttpEntity httpEntity = new UrlEncodedFormEntity(nvps);
ByteBuf byteBuf = 
Unpooled.copiedBuffer(EntityUtils.toByteArray(httpEntity));
request.content().writeBytes(byteBuf);
request.headers().set(HttpHeaders.Names.CONTENT_LENGTH,request.content().readableBytes());
fu.channel().writeAndFlush(request)
于 2018-01-05T06:52:55.880 回答