小编Mat*_*bst的帖子

我想对我自己编写的HTTP Servlet发出POST请求.良好的情况(HTTP响应代码200)始终使用URL.openConnection()方法正常工作.但是当我收到所需的错误响应代码(例如400)时,我想我必须使用HttpUrlConnection.getErrorStream().但ErrorStream对象为null虽然我在错误情况下从servlet发回数据(我想评估此数据以生成错误消息).这就是我的代码:

HttpURLConnection con = null;
        try {
            //Generating request String
            String request = "request="+URLEncoder.encode(xmlGenerator.getStringFromDocument(xmlGenerator.generateConnectRequest(1234)),"UTF-8");
            //Receiving HttpUrlConnection (DoOutput = true; RequestMethod is set to "POST")
            con = openConnection();
            if (con != null){
                PrintWriter pw = new PrintWriter(con.getOutputStream());
                pw.println(request);
                pw.flush();
                pw.close();
                InputStream errorstream = con.getErrorStream();

                BufferedReader br = null;
                if (errorstream == null){
                    InputStream inputstream = con.getInputStream();
                    br = new BufferedReader(new InputStreamReader(inputstream));
                }else{
                    br = new BufferedReader(new InputStreamReader(errorstream));
                }
                String response = "";
                String nachricht;
                while ((nachricht = br.readLine()) != null){ …