嗨,有人可以给我一个示例代码,每五分钟获取一次位置,请尝试一下,我可以通过按钮按钮获取位置, 但我需要它显示一次五分钟。
谢谢
这是我的代码:
public void checkLocation(View v) {
//initialize location manager
manager = (LocationManager) getSystemService(Context.LOCATION_SERVICE);
//check if GPS is enabled
//if not, notify user with a toast
if (!manager.isProviderEnabled(LocationManager.GPS_PROVIDER)) {
Toast.makeText(this, "GPS is disabled.", Toast.LENGTH_SHORT).show();
} else {
//get a location provider from location manager
//empty criteria searches through all providers and returns the best one
String providerName = manager.getBestProvider(new Criteria(), true);
Location location = manager.getLastKnownLocation(providerName);
TextView tv = (TextView)findViewById(R.id.locationResults);
if (location != null) {
tv.setText(location.getLatitude() + " latitude, " + location.getLongitude() + " longitude");
} else {
tv.setText("Last known location not found. Waiting for updated location...");
}
//sign up to be notified of location updates every 15 seconds - for production code this should be at least a minute
manager.requestLocationUpdates(providerName, 15000, 1, this);
}
}
@Override
public void onLocationChanged(Location location) {
TextView tv = (TextView)findViewById(R.id.locationResults);
if (location != null) {
tv.setText(location.getLatitude() + " latitude, " + location.getLongitude() + " longitude");
} else {
tv.setText("Problem getting location");
}
}
@Override
public void onProviderDisabled(String arg0) {}
@Override
public void onProviderEnabled(String arg0) {}
@Override
public void onStatusChanged(String arg0, int arg1, Bundle arg2) {}
// Find the closest Bart Station
public String findClosestBart(Location loc) {
double lat = loc.getLatitude();
double lon = loc.getLongitude();
double curStatLat = 0;
double curStatLon = 0;
double shortestDistSoFar = Double.POSITIVE_INFINITY;
double curDist;
String curStat = null;
String closestStat = null;
//sort through all the stations
// write some sort of for loop using the API.
curDist = Math.sqrt( ((lat - curStatLat) * (lat - curStatLat)) +
((lon - curStatLon) * (lon - curStatLon)) );
if (curDist < shortestDistSoFar) {
closestStat = curStat;
}
return closestStat;
}
答案 0 :(得分:11)
以下是获取位置的代码,并设置gps的侦听器在几分钟和距离上获取当前位置,我也使用runnable对象每隔几分钟获取一次位置。
Location gpslocation = null;
private static final int GPS_TIME_INTERVAL = 60000; // get gps location every 1 min
private static final int GPS_DISTANCE= 1000; // set the distance value in meter
/*
for frequently getting current position then above object value set to 0 for both you will get continues location but it drown the battery
*/
private void obtainLocation(){
if(locMan==null)
locMan = (LocationManager) getSystemService(LOCATION_SERVICE);
if(locMan.isProviderEnabled(LocationManager.GPS_PROVIDER)){
gpslocation = locMan.getLastKnownLocation(LocationManager.GPS_PROVIDER);
if(isLocationListener){
locMan.requestLocationUpdates(LocationManager.GPS_PROVIDER,
GPS_TIME_INTERVAL, GPS_DISTANCE, GPSListener);
}
}
}
}
现在使用此方法获取当前位置,并且每隔1分钟和1000米距离调用位置更改的监听器。
为了每5分钟获得一次,您可以使用此处理程序并运行以在设定良好的时间段内获取此位置:
private static final int HANDLER_DELAY = 1000*60*5;
Handler handler = new Handler();
handler.postDelayed(new Runnable() {
public void run() {
myLocation = obtainLocation();
handler.postDelayed(this, HANDLER_DELAY);
}
}, START_HANDLER_DELAY);
以下是位置变更事件的GPS监听器:
private LocationListener GPSListener = new LocationListener(){
public void onLocationChanged(Location location) {
// update location
locMan.removeUpdates(GPSListener); // remove this listener
}
public void onProviderDisabled(String provider) {
}
public void onProviderEnabled(String provider) {
}
public void onStatusChanged(String provider, int status, Bundle extras) {
}
};
您可以为侦听器和处理程序设置相同的间隔时间以获取GPS位置。
答案 1 :(得分:2)
嗨使用以下计时器代码。
您可以使用以下选项 选项1 如果移动设备移动100米,这将获得位置。
captureFrequencey=3*60*1000;
LocationMngr.requestLocationUpdates(LocationManager.GPS_PROVIDER, captureFrequencey, 100, this);
选项2
TimerTask refresher;
// Initialization code in onCreate or similar:
timer = new Timer();
refresher = new TimerTask() {
public void run() {
handler.sendEmptyMessage(0);
};
};
// first event immediately, following after 1 seconds each
timer.scheduleAtFixedRate(refresher, 0,1000);
//=======================================================
final Handler handler = new Handler() {
public void handleMessage(Message msg) {
switch (msg.what) {
case REFRESH:
//your code here
break;
default:
break;
}
}
};
计时器将调用处理程序持续一段时间(将1000更改为您所需的时间)。
希望这会对你有所帮助。
答案 2 :(得分:1)
我使用runnable来做这件事,
final Runnable r = new Runnable() {
public void run() {
//Here add your code location listener call
handler.postDelayed(this, 300000 );
}
};
handler.postDelayed(r, 300000 );
答案 3 :(得分:0)
尝试这样:
private Handler handler = new Handler();
handler.postDelayed(runnable, 300000);
private Runnable runnable = new Runnable() {
public void run() {
if (location != null) {
onLocationChanged(location);
} else {
System.out.println("Location not avilable");
}
handler.postDelayed(this, 300000);
}
};