如何讀取HTTP POST請求返回的XML？

我送一個POST要求是這樣的：

 String urlParameters = "a=b&c=d"; 
     String request = "http://www.example.com/"; 

     URL url = new URL(request); 
     HttpURLConnection connection = (HttpURLConnection) url.openConnection();  
     connection.setDoOutput(true); 
     connection.setDoInput(true); 
     connection.setInstanceFollowRedirects(false); 
     connection.setRequestMethod("POST"); 
     connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded"); 
     connection.setRequestProperty("charset", "utf-8"); 
     connection.setRequestProperty("Content-Length", "" + Integer.toString(urlParameters.getBytes().length)); 
     connection.setUseCaches(false); 

     DataOutputStream wr = new DataOutputStream(connection.getOutputStream()); 
     wr.writeBytes(urlParameters); 
     wr.flush(); 
     wr.close(); 
     connection.disconnect();

我如何讀取從HTTP POST請求返回的XML響應？特別是，我想將響應文件保存爲.xml文件，然後讀取它。對於我平時GET請求，我用這個：

SAXBuilder builder = new SAXBuilder(); 
    URL website = new URL(urlToParse); 
    ReadableByteChannel rbc = Channels.newChannel(website.openStream()); 
    FileOutputStream fos = new FileOutputStream("request.xml"); 
    fos.getChannel().transferFrom(rbc, 0, 1 << 24); 
    fos.close(); 
    // Do the work

附錄：我使用下面的代碼和它工作得很好。但是，它會忽略任何空格和換行符，並將完整的XML內容視爲單行。我如何解決它？

InputStream is = connection.getInputStream(); 
    BufferedReader br = new BufferedReader(new InputStreamReader(is)); 
    StringBuilder sb1 = new StringBuilder(); 
    String line; 
    while ((line = br.readLine()) != null) { 
     sb1.append(line); 
    } 
    FileOutputStream f = new FileOutputStream("request.xml"); 
    f.write(sb1.toString().getBytes()); 
    f.close(); 
    br.close();

來源

2013-04-08 Karan Goel

嘗試從連接獲取的InputStream從中讀取，你寫完你的OutputStream後，對不對？ – bdares 2013-04-08 23:58:53

@bdares請參閱OP進行編輯。 – 2013-04-09 00:45:36

不使用Readers和readLine() XML數據。使用InputStreams和byte[] s。

來源

2013-04-09 00:55:36 jtahlborn

感謝盤古，我修改了他的代碼，這現在工作：

TransformerFactory transFactory = TransformerFactory.newInstance(); 
    Transformer t= transFactory.newTransformer(); 
    t.setOutputProperty(OutputKeys.METHOD, "xml"); 
    t.setOutputProperty(OutputKeys.INDENT,"yes");     
    Source input = new StreamSource(is); 
    Result output = new StreamResult(new FileOutputStream("request.xml")); 
    transFactory.newTransformer().transform(input, output);

來源

2013-04-09 01:12:49

如何讀取HTTP POST請求返回的XML？

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