location

How can I get location without internet in android, using only GPS

匿名 (未验证) 提交于 2019-12-03 02:16:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I want to get location using GPS only. I don't want to use internet and GPRS in this application. My code is below; tell me where I'm wrong in this. code: package com . getlocation ; import android . app . Activity ; import android . content . Context ; import android . location . Criteria ; import android . location . Location ; import android . location . LocationListener ; import android . location . LocationManager ; import android . os . Bundle ; import android . util . Log ; import android . widget . Toast ; public class

Pandas: resample timeseries with groupby

匿名 (未验证) 提交于 2019-12-03 02:14:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: Given the below pandas DataFrame: In [115]: times = pd.to_datetime(pd.Series(['2014-08-25 21:00:00','2014-08-25 21:04:00', '2014-08-25 22:07:00','2014-08-25 22:09:00'])) locations = ['HK', 'LDN', 'LDN', 'LDN'] event = ['foo', 'bar', 'baz', 'qux'] df = pd.DataFrame({'Location': locations, 'Event': event}, index=times) df Out[115]: Event Location 2014-08-25 21:00:00 foo HK 2014-08-25 21:04:00 bar LDN 2014-08-25 22:07:00 baz LDN 2014-08-25 22:09:00 qux LDN I would like resample the data to aggregate it hourly by count while grouping by location

Android - Unable to get the gps location on the emulator

匿名 (未验证) 提交于 2019-12-03 02:14:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: i'm tring to use the gps on the android emulator, i've the following code: public class NL extends Activity { private LocationManager locmgr = null; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.nl); locmgr = (LocationManager) getSystemService(Context.LOCATION_SERVICE); Criteria crit = new Criteria(); crit.setAccuracy(Criteria.ACCURACY_FINE); String provider = locmgr.getBestProvider(crit, true); Location loc = locmgr.getLastKnownLocation(provider); Toast msg = Toast

how to get current location in google map android

匿名 (未验证) 提交于 2019-12-03 02:11:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: Actually my problem is I am not getting current location latitude and longitude I tried so many ways.I know that this question already asked in SO I tried that answers also still I didn't get answer.Please help me Code: if (googleMap == null) { googleMap = ((MapFragment) getFragmentManager().findFragmentById( R.id.map)).getMap(); // check if map is created successfully or not if (googleMap == null) { Toast.makeText(getApplicationContext(), "Sorry! unable to create maps", Toast.LENGTH_SHORT) .show(); } } googleMap.setMyLocationEnabled(true);

NETWORK_PROVIDER not providing updated locations

匿名 (未验证) 提交于 2019-12-03 02:11:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: LocationManager locationManager = (LocationManager)getApp().getSystemService(Context.LOCATION_SERVICE); //getApp() returns my Application object locationManager.requestLocationUpdates(LocationManager.GPS_PROVIDER , 1, 1, this); locationManager.requestLocationUpdates(LocationManager.NETWORK_PROVIDER, 1, 1, this); That's the code I'm using to listen. With GPS enabled, everything works fine. However, if I disable GPS and rely on network location, it gives me stale results -- in this case, from two days ago. I cannot get it to update. Calling

How to find all connected components in a binary image in Matlab?

匿名 (未验证) 提交于 2019-12-03 02:11:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I have been trying to find all connected components using 8 neighbors in a binary image, without using the function "bwlabel". For example, my input matrix is: a = 1 1 0 0 0 0 0 1 1 0 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 I would to have something like this: a = 1 1 0 0 0 0 0 1 1 0 0 2 2 0 1 1 0 0 0 2 0 1 1 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 There are 3 connected objects in this image. 回答1: This is a common problem in image processing. There are many variations, such as flood filling a region in an

Bootstrap 3: Keep selected tab on page refresh

匿名 (未验证) 提交于 2019-12-03 02:11:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I am trying to keep selected tab active on refresh with Bootstrap 3 . Tried and checked with some question already been asked here but none of work for me. Don't know where I am wrong. Here is my code HTML <!-- tabs link --> <ul class = "nav nav-tabs" id = "rowTab" > <li class = "active" ><a href = "#personal-info" data-toggle = "tab" > Personal Information </a></li> <li><a href = "#Employment-info" data-toggle = "tab" > Employment Information </a></li> <li><a href = "#career-path" data-toggle = "tab" > Career Path </a></li> <li><a

用nginx解决前端跨域问题

喜你入骨 提交于 2019-12-03 02:10:54
假如前端你项目部署在nginx的根目录下,然后项目需要请求后台小伙伴写的接口 nginx配置: #user nobody; worker_processes 1; #error_log logs/error.log; #error_log logs/error.log notice; #error_log logs/error.log info; #pid logs/nginx.pid; events { worker_connections 1024; } http { include mime.types; default_type application/octet-stream; #log_format main '$remote_addr - $remote_user [$time_local] "$request" ' # '$status $body_bytes_sent "$http_referer" ' # '"$http_user_agent" "$http_x_forwarded_for"'; #access_log logs/access.log main; sendfile on; #tcp_nopush on; #keepalive_timeout 0; keepalive_timeout 65; #gzip on; server { listen

How to dereference a memory location from python ctypes?

匿名 (未验证) 提交于 2019-12-03 02:06:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I want to replicate the following c code in python ctypes: main() { long *ptr = (long *)0x7fff96000000; printf("%lx",*ptr); } I can figure out how to call this memory location as a function pointer but not just do a normal dereference: from ctypes import * """ >>> fptr = CFUNCTYPE(None, None) Traceback (most recent call last): File " ", line 1, in File "/usr/lib/python2.6/ctypes/__init__.py", line 104, in CFUNCTYPE class CFunctionType(_CFuncPtr): TypeError: Error when calling the metaclass bases item 1 in _argtypes_ has no from_param method

Service not Available - Geocoder Android [duplicate]

匿名 (未验证) 提交于 2019-12-03 02:06:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: This question already has an answer here: Why is Android Geocoder throwing a “Service not Available” exception? 5 answers I have a little problem with the geocoder to get latitude and lontitude from an address : This is my code: for (Garage g : XMLGarage) { List address; address = coder.getFromLocationName( g.getAddress(), 5); if (address != null) { Address location = address.get(0); g.setLatitude(location.getLatitude()); g.setLongitude(location.getLongitude()); ... The error is : 08-18 14:28:56.026: WARN/System.err(359): java.io.IOException