gps

Cordova Angularjs ask for GPS

南楼画角 提交于 2019-12-06 15:09:18
I program an app where I need GPS. If the GPS is off, how can I ask the user to turn on GPS, like google maps? After a while of googleing I found nothing compareable... My other question is, why I get "Use location? - 2", when the GPS is off? I thought, if the GPS is off, $window.navigator.geolocation is false? controller.js if ($window.navigator && $window.navigator.geolocation) { function success(pos) { $rootScope.position = { x: pos.coords.latitude, y: pos.coords.longitude }; } function fail(error) { alert("Use location? - 2"); } $window.navigator.geolocation.getCurrentPosition(success,

Converting GPS coordinates (latitude and longitude) to decimal

笑着哭i 提交于 2019-12-06 14:26:12
I am getting the latitude and longitude from GPS device. They look like 21081686N,079030977E If I manually convert them to 21°08'16.86"N, 79°03'09.77"E and check on Google maps, the location is almost correct. How do I convert these values in java and convert them to decimal accurately? Any JAVA libraries? Thanks, acraig5075 You shouldn't need a library, but rather just break up the strings into their component parts. Perhaps this will help. Please note I'm not a Java programmer so this syntax could very well be wrong. (Perhaps retag your question to Java) By decimal I presume you mean decimal

Altitude problem IOS

末鹿安然 提交于 2019-12-06 14:17:56
问题 I have a problem in getting the altitude , it returns 0.0000 . Here is my code: - (void)locationManager:(CLLocationManager *)manager didUpdateToLocation:(CLLocation *)newLocation fromLocation:(CLLocation *)oldLocation { NSString *tAltitude = [NSString stringWithFormat:@"%f", [newLocation altitude]]; CLLocationCoordinate2D coord=newLocation.coordinate; MKCoordinateSpan span = {.latitudeDelta = 0.005, .longitudeDelta = 0.005}; MKCoordinateRegion region = {coord, span}; [map setRegion:region];

高通区块狗模式现成APP出售

孤街醉人 提交于 2019-12-06 13:51:18
高通区块狗系统 APP开发附源码, ( 135-3970-7842 陈经理微/电同号) 高通区块狗系统开发,高通区块狗 APP专业开发,高通区块狗系统软件定制,高通区块狗模式现成APP出售 大家应该都知道这个 GPS 的系统它是美国人研发出来的技术,而且它已经在市场上出现好多年了。一说美国都懂,它是一个发达国家,不管是经济还是国力上都比咱们中国要好太多。特别是他们的科学技术水平在世界上可以说是数一数二的水准。很多人用这个 GPS 也已经用习惯,一下子换成别的还真的可能觉得没有原来的好用。 ----注!!!专业开发公司,非平台运营方,玩家勿扰!!!----- 一、预约领 . 养 1 代区块狗    1. 活动期间内,用户点击页面下方“立即预约”按钮,填写用户信息,成功提交区块狗预约领 . 养登记后,即可领 . 养 1 代区块狗 1 只。    2. 每位用户 ( 同一微 * 信号、手机号、设备、 IP 地址视为同一用户 ) ,可通过预约方式,领 . 养最多 1 只 1 代区块狗。    3. 每位用户,只能预约 1 次,重复预约无效。 二、转发本活动领 . 养 0 代区块狗    1. 活动期间内,用户将本活动成功分享至微 *信好友或微*信朋友圈,邀请微* 信好友点击页面中 “帮 TA 实现愿望”按钮,集齐 10 个微信好友的帮助,即可领 . 养 0 代区块狗 1 只 135 可微

android onLocationChanged never called

被刻印的时光 ゝ 提交于 2019-12-06 13:50:29
问题 I've implemented code that should return present location. First the code: public static double[] getLocation(Context context) { double[] result; lm = (LocationManager)context.getSystemService(Context.LOCATION_SERVICE); MyLocationListener ll = new MyLocationListener(); lm.requestLocationUpdates(LocationManager.GPS_PROVIDER, 0, 0, ll); while(!hasLocation) { } result = ll.getResult(); return result; } private static class MyLocationListener implements LocationListener { double[] result = new

Pinpointing indoor location with android - not accurate enough

你。 提交于 2019-12-06 13:34:40
I'm trying to find a solution for pinpointing indoors, specifically inside big crowded places, like malls for example. I tried building a simple app that tried to pinpoint the phone using only the Network service or Wifi networks. Usually when Wifi was on the accuracy got to 40-60 meters, and when the wifi was off and only the network service was used, the accuracy got to 600-1000 meters. Unfortunately that really is too inaccurate for my needs. I read that Google have successfully added an accurate solution for indoors navigation ( http://www.engadget.com/2012/01/08/google-maps-indoor

Kalman-filtered GPS data is still fluctuating a lot

风流意气都作罢 提交于 2019-12-06 13:20:42
He everyone! I'm writing an Android app that uses the devices GPS to calculate a vehicles driving speed. This is supposed to be accurate to about 1-2 km/h, and I'm doing it by looking at the distance between two GPS locations and dividing it by the time these locations are apart, pretty straightforward, and then doing this for the last three recorded coordinates and evening it out. I get the GPS data in a background service, that has a handler to it's own looper, so whenever I get a new location from the LocationListener, I call the Kalmans update() method and call the predict() in a handler

How to find the gps in the android device is correctly working or not?

荒凉一梦 提交于 2019-12-06 12:47:25
问题 I'm developing an android application for mobiles and tablets. I'm using android version 2.2 , API 8. In my application, I want to capture the location co-ordinates via gps and send to server. It's working fine . When I working in my office (near to my office) the gps - co-ordinates captured correctly in android devices But in only one device the co-ordinates captured wrongly. I'm in Alwarpet,chennai,india .But, I run that device it shows " Andra pradesh" or bay of "bengal ".But all other

Android Location not returning Location from Custom Class

岁酱吖の 提交于 2019-12-06 12:38:49
问题 I have my own location class. I have an odds result that when I search for the GPS location using my class, I get 0, 0 back. Whereas if I search the getLastKnown function in location manager for GPS, I get the forced value: public Location getGPSloc(Context c){ isGPSavailable(c); if(gps_enabled) lm.requestLocationUpdates(LocationManager.GPS_PROVIDER, 5000, 0, locationListenerGps); timerGPS = new Timer(); timerGPS.schedule(new GetLastLocationGPS(), 45000); return gpsLoc; } This is the class

How to visualise an arrow pointing to a certain coordinates(latitude, longitude), given pich and yaw

霸气de小男生 提交于 2019-12-06 12:37:21
问题 I would like to know how to visualise an arrow pointing to a certain coordinates(latitude, longitude) while I'm moving around holding my handset. I'va calculated the pitch and yaw of the desired position. Now I want to know how to keep pointing to this position while moving. Regards, 回答1: In general you can calculate the bearing angle (the angle between your local meridian and the great circle connecting your current position and the target position measured from the north direction) using