7

嗨，任何人都可以给我一个示例代码，以每五分钟获取一次位置，请我尝试过，我可以通过单击按钮获取一次位置，但我需要它显示一次五分钟。

谢谢你

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;

        }
4

4 回答 4

13

这是获取位置的代码并设置gps的侦听器以在几分钟和距离内获取当前位置，我也使用可运行对象每隔几分钟获取位置。

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，并在每 1 分钟和 1000 米距离发生位置变化时调用侦听器。

为了每 5 分钟获取一次，您可以使用此处理程序和 runnable 在设定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);

这是位置更改事件的 GPS 侦听器：

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) {
    }
};

您可以为获取 GPS 位置的侦听器和处理程序设置相同的间隔时间。

于 2013-04-17T08:55:45.040 回答
2

嗨使用下面的计时器代码。

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);

看看这个链接http://developer.android.com/reference/android/location/LocationManager.html#requestLocationUpdates%28java.lang.String,%20long,%20float,%20android.location.LocationListener%29

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;
              }
          }
        };

Timer 将在您的持续时间内调用处理程序（将 1000 更改为您需要的时间）。

希望这会帮助你。

于 2013-04-17T08:52:07.157 回答
1

我用 runnable 来做这个，

    final Runnable r = new Runnable() {
        public void run() {
    //Here add your code location listener call
    handler.postDelayed(this, 300000 );
        }
    };

    handler.postDelayed(r, 300000 );
于 2013-04-17T08:50:46.170 回答
0

试试这样：

 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);
    } 
};
于 2013-04-17T08:53:16.973 回答