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数据抛出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 ？

Answer 1

  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); } 

Answer 2

使用Apache HTTP Compennts 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。

Answer 3

特别是Apache HTTP Components ， 客户端将是最好的方式。 它吸收了很多你通常不得不亲手做的令人讨厌的编码