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 读取它并将其插入到 MySql 对象中。我已经使用这个 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 的帮助下完成此操作。 早些时候我通过如下传递 url 参数来做到这一点并且它工作正常,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://stackoverflow.com/questions/11074934/the-json-object-sent-from-android-application-is-null-when-i-want-to-access-他)但听不懂。
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;
}
最佳答案
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);
}
关于java - 如何使用 doGet 方法捕获从 android 应用程序发送到 servlet 的 JSON 对象?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11242868/