如何通过url.openStream（）发送POST数据？
我正在寻找教程或快速示例，我如何发送POST数据扔openStream。
我的代码是：
URL url = new URL("http://localhost:8080/test");
InputStream response = url.openStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(response, "UTF-8"));
你可以帮帮我吗 ？
URL url = new URL(urlSpec);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod(method);
connection.setDoOutput(true);
connection.setDoInput(true);
// important: get output stream before input stream
OutputStream out = connection.getOutputStream();
out.write(content);
out.close();
// now you can get input stream and read.
BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String line = null;
while ((line = reader.readLine()) != null) {
writer.println(line);
}
使用Apache HTTP Compoennts http://hc.apache.org/httpcomponents-client-ga/
教程：http://hc.apache.org/httpcomponents-client-ga/tutorial/html/fundamentals.html
寻找HttpPost - 有一些发送动态数据，文本，文件和表单数据的例子。
尤其是Apache HTTP组件，客户端将是最好的选择。 它可以避免许多你通常需要手工完成的令人讨厌的编码
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