Android无需按下按钮即可将Gps数据发送到服务器
Android无需按下按钮即可将Gps数据发送到服务器
提问于 2014-07-31 21:23:50
POST服务目前正在后台运行，有什么建议吗？？我必须实现一个计时器吗?不知道它超出了我的领域吗？提前感谢!！
Button btnCreateUser = (Button) findViewById(R.id.btnSend);
// button click event
btnCreateUser.setOnClickListener(new View.OnClickListener() {
public void onClick(View view) {
// creating new product in background thread
new CreateNewUser().execute();
}
});
}
class CreateNewUser extends AsyncTask<String, String, String> {
@Override
protected String doInBackground(String... args) {
String lat=latitude, lng=longitude;
SimpleDateFormat sdfDateTime = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss", Locale.US);
String newtime = sdfDateTime.format(new Date(System.currentTimeMillis()));
EditText username = (EditText) findViewById(R.id.myText);
String newString= (String)username.getText().toString();
String created_at=newtime;
String timezone="UTC";
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url_create_user);
String responseBody = "";
HttpResponse response = null;
try {
String base64EncodedCredentials = "Basic " + Base64.encodeToString(
(username + ":" + created_at +":"+timezone+":"+lat+":"+lng).getBytes(),
Base64.NO_WRAP);
httppost.setHeader("Authorization", base64EncodedCredentials);
httppost.setHeader(HTTP.CONTENT_TYPE,"application/json");
JSONObject obj = new JSONObject();
obj.put("username", String.valueOf(newString));
obj.put("created_at",String.valueOf(newtime));
obj.put("timezone", String.valueOf(timezone));
obj.put("lat", String.valueOf(latitude));
obj.put("lng", String.valueOf(longitude));
httppost.setEntity(new StringEntity(obj.toString(), "UTF-8"));
// Execute HTTP Post Request
response = httpclient.execute(httppost);
if (response.getStatusLine().getStatusCode() == 200) {
Log.d("response ok", "Reached Server ok");
startActivity(new Intent("com.xxxxx.CLEARSCREEN"));
} else {
Log.d("response not ok", "Something went wrong :/");
}
responseBody = EntityUtils.toString(response.getEntity());
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
catch (JSONException e) {
e.printStackTrace();
}
return responseBody;
}回答 2
Stack Overflow用户
发布于 2014-07-31 21:40:36
你可以通过以下代码来解决这个问题：
Timer timer = new Timer();
timer.scheduleAtFixedRate(new TimerTask() {
public void run() {
btnCreateUser.post(new Runnable() {
@Override
public void run() {
btnCreateUser .performClick();
}
});
}
}, 0, 14400); // for 4 hoursStack Overflow用户
发布于 2014-08-06 18:59:45
使用AlarmManager (http://developer.android.com/reference/android/app/AlarmManager.html)在将来重复启动您的活动。
假设您用于收集用户数据并将其发送到服务器的活动名为YourActivity，只需添加一些静态方法即可，如下所示，用于安排活动启动。您可以在应用程序启动时或从BOOT_COMPLETE BroadcastReceiver调用schedule方法。
public static void schedule(Context context, String url,
long triggerAtMillis, long intervalMillis) {
final AlarmManager alarmManager =
(AlarmManager) context.getSystemService(ALARM_SERVICE);
alarmManager.setRepeating(AlarmManager.RTC, triggerAtMillis,
intervalMillis, createPendingOperation(context));
}
public static void unschedule(Context context) {
final AlarmManager alarmManager =
(AlarmManager) context.getSystemService(ALARM_SERVICE);
alarmManager.cancel(createPendingOperation(context));
}
private static PendingIntent createPendingOperation(Context context) {
return PendingIntent.getActivity(
context, 0, new Intent(context, YourActivity.class),
PendingIntent.FLAG_CANCEL_CURRENT);
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/25060024
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