如何使用doGet方法捕获从android应用程序发送到servlet的JSON对象

Question

我想从用户那里获取用户名，密码和电子邮件ID，构造一个JSON对象并将其发送到java servlet，然后Java servlet读取它并将其插入MySql对象中。我已经用了一个php服务器（来源：http://www.androidhive.info/2011/10/android-login-and-registration-screen-design/），但我需要在java servlet的帮助下做到这一点。早些时候我通过按如下方式传递 url 参数来执行此操作，它工作正常，但现在我想将这些信息用作 JSON 参数：

安卓代码：

try {
                url = new URL("http://10.0.2.2:8080/Servlet/Servlet?param1="
                        + uname + "&param2=" + pwd + "&param3=" + eid);
                // url = new URL("http://10.0.2.2:8080/Servlet/Servlet");
                HttpURLConnection urlConnection = (HttpURLConnection) url
                        .openConnection();
                InputStream in = new BufferedInputStream(
                        urlConnection.getInputStream());
                urlConnection.disconnect();
            } catch (Exception e) {
                e.printStackTrace();
            }

Servlet Code：

protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException
    {
        req.setCharacterEncoding("UTF-8");
        resp.setCharacterEncoding("UTF-8");
        final String uname = req.getParameter("param1");
        final String pwd = req.getParameter("param2");
        final String eid = req.getParameter("param3");

我看了一下这个（http://stackoverflow.com/questions/11074934/the-json-object-sent-from-android-application-is-null-when-i-want-to-access-him），但无法理解。

JSON代码如下（来源：http://www.androidhive.info/2011/10/android-login-and-registration-screen-design/）：

public class JSONParser {
    static InputStream is = null;
    static JSONObject jObj = null;
    static String json = "";
    // constructor
    public JSONParser() {
    }
    public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) {
        // Making HTTP request
        try {
            // defaultHttpClient
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);
            httpPost.setEntity(new UrlEncodedFormEntity(params));
            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "n");
            }
            is.close();
            json = sb.toString();
            Log.e("JSON", json);
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }
        // try parse the string to a JSON object
        try {
            jObj = new JSONObject(json);            
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }
        // return JSON String
        return jObj;
    }
}

public JSONObject loginUser(String email, String password){
        // Building Parameters
        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("tag", login_tag));
        params.add(new BasicNameValuePair("email", email));
        params.add(new BasicNameValuePair("password", password));
        JSONObject json = jsonParser.getJSONFromUrl(loginURL, params);
        // return json
        // Log.e("JSON", json.toString());
        return json;
    }

Answer 1

首先，您创建一个 Post 请求。 通常建议从 Get 调用帖子。

protected void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException
{
    doPost(request, response);
}

这就是我为构建他请求的 JSON 对象所做的

 */
protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException
{
    HttpSession httpSession = request.getSession(false);
    JSONObject jJobObject = JSONObject.fromObject(request.getParameter("data"));
    JJob jJob = (JJob) JSONObject.toBean(jJobObject, JJob.class);
    String strTerm = (String) httpSession.getAttribute("terminal");
    Integer term = null;
    try {
        term = Integer.parseInt(strTerm);
    }
    catch(Exception e) {
        //
    }
    jJob = PersoJobService.createJob(jJob, (Integer) httpSession.getAttribute("userId"), term );
    writeResponse(JSONObject.fromObject(jJob), request, response);
}

Answer 2

你可以使用简单的序列化，方法是编写准备好的 JSONObject 对象并将其序列化到 servlet。如果你从java到java进行通信，序列化非常方便。

Map<Object, Object> data = new Hashtable<Object, Object>(0);
data.put("etc", "etc");
...
URLConnection con = url.openConnection();
con.setDoInput(true);
con.setDoOutput(true);
new ObjectOutputStream(con.getOutputStream()).writeObject(data);

并从servlet

Map<Object, Object> data = (Map<Object, Object>) new ObjectInputStream(request.getInputStream()).readObject();

您甚至可以通过此方法传递复杂的可序列化对象。您可以从服务写入对象并在客户端读取。