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，然后读取它并将其插入 MySql 对象。我已经使用了这个 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。 早些时候我通过如下传递 url 参数来做到这一点，它工作正常，但现在我想将信息用作 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 代码：

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");

我看过这个（http://stackoverflow.com/questions/11074934/the-json-object-sent-from-android-application-is-null-when-i-want-to-access-他）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，您要创建一个 Post 请求。一般建议从get调用post。

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

这就是我根据他的请求构建 JSON 对象的方法

 */
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

您可以通过编写准备好的 JSONObject 对象并将其序列化到 servlet 来使用简单的序列化。如果您从 java 到 java 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);

并且来自 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。 您也可以从服务写入对象并在客户端读取。