如何使用 doGet 方法捕获从 android 应用程序发送到 servlet 的 JSON 对象?
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【中文标题】如何使用 doGet 方法捕获从 android 应用程序发送到 servlet 的 JSON 对象?【英文标题】:How can I catch a JSON object sent from an android app to a servlet with doGet method? 【发布时间】:2012-06-29 20:50:35 【问题描述】: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 + "¶m2=" + pwd + "¶m3=" + 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://***.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;
【问题讨论】:
那么这个代码中的问题....? 问题是我如何在我的 servlet 中捕获 JSONrequest,就像我正在捕获 url 参数一样? 好的..你不能在单个参数中发布完整的 json 字符串吗? see this 更好的解决方案是将其作为发布请求发送。 【参考方案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);
【讨论】:
【参考方案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。
【讨论】:
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