我的Android项目提供基于位置的提醒。在服务中,我正在计算从当前位置到数据库中所有位置的距离。如果距离小于500米,则向用户发出通知。问题在于,由于距离的计算和通知的发送正在服务中进行,如果距离<500m,则通知会在相同位置重复进行。最终应用程序必须强行关闭。这是服务类的onLocationChanged方法:
public void onLocationChanged(Location location) {
double lat = location.getLatitude(),lng = location.getLongitude(),dblat = 0,dblng=0;
Location currentlocation = new Location("current Location");
currentlocation.setLatitude(lat);
currentlocation.setLongitude(lng);
int NOTIFICATION_ID = 1;
String ns = Context.NOTIFICATION_SERVICE;
String title,date,today;
NotificationManager mNotificationManager = (NotificationManager) getSystemService(ns);
Calendar cal=Calendar.getInstance();
today = String.valueOf(cal.get(Calendar.YEAR)) + "-"
+ String.valueOf(cal.get(Calendar.MONTH)) + "-"
+ String.valueOf(cal.get(Calendar.DAY_OF_MONTH));
SQLiteDatabase database = new SQLiteHelper(getApplicationContext()).getWritableDatabase();
Cursor c = database.rawQuery("select title,reminderdate,latitude,longitude from Reminder where type=1", null);
int icon = R.drawable.ic_launcher;
if(c.getCount()>0)
{ //System.out.println("c not null");
c.moveToFirst();
while(!c.isAfterLast())
{
title=c.getString(0);
date=c.getString(1);
if(date.equals(today))
{
dblat=c.getDouble(2);
dblng=c.getDouble(3);
Location dblocation = new Location("db Location");
dblocation.setLatitude(dblat);
dblocation.setLongitude(dblng);
Double distance = (double) currentlocation.distanceTo(dblocation);
if(distance<500)
//Toast.makeText(this, "distance " + distance, Toast.LENGTH_LONG).show();
{ long when = System.currentTimeMillis();
Notification notification = new Notification(icon,title, when);
RemoteViews contentView = new RemoteViews(getPackageName(), R.layout.custom_notification);
contentView.setImageViewResource(R.id.notification_image, R.drawable.ic_launcher);
contentView.setTextViewText(R.id.notification_title, "My custom notification title");
contentView.setTextViewText(R.id.notification_text, "My custom notification text");
notification.contentView = contentView;
Intent notificationIntent = new Intent(this, ShowReminder.class);
notificationIntent.putExtra("title", title);
PendingIntent contentIntent = PendingIntent.getActivity(this, 0, notificationIntent, 0);
notification.contentIntent = contentIntent;
notification.flags |= Notification.FLAG_NO_CLEAR; //Do not clear the notification
notification.defaults |= Notification.DEFAULT_LIGHTS; // LED
notification.defaults |= Notification.DEFAULT_VIBRATE; //Vibration
notification.defaults |= Notification.DEFAULT_SOUND; // Sound
mNotificationManager.notify(NOTIFICATION_ID, notification);
} }
c.moveToNext();
}}
c.close();
database.close();
}
每当我靠近它而不是反复出现时,我怎样才能为特定位置只发出一次通知?任何帮助将不胜感激。非常感谢!
答案 0 :(得分:1)
您必须为服务添加一些逻辑,以便在触发通知时更加谨慎地决定。
例如,您可以向表中添加一列,用于跟踪用户当前是否“足够接近”(在您的示例中为<500)到要通知的给定位置;每次重新计算距离时,如果您发现自己现在“足够接近”但该列尚未反映出来,那么您知道用户刚刚进入该位置附近并且是时候将通知排入队列了关于它。在数据库中记下用户现在靠近该位置,以便您可以抑制有关该位置的冗余通知。一旦离开附近,请务必清除“足够靠近”位。
如果您处于“足够接近”的边缘,仍然会有一些冗余通知,因此您需要添加一些去抖动,但这应该可以让您继续前进。