如何通过url.openStream（）发送POST数据？

Question

我正在寻找教程或快速的例子，我怎么可以发送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")); 

你能帮助我吗？

Answer 1

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

查找HttpPost - 有发送动态数据，文本，文件和表单数据的一些例子。

Answer 2

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 3

Apache HTTP Components特别是Client将是最好的一段路要走。 它大量的令人讨厌的编码，你通常必须做手