Android 移除并重新插入GeoFire位置,以触发onKeyExited和onKeyEntered不';行不通
我对GeoFire和Firebase有问题 我想根据GeoFire位置触发器更新UI上的信息。当客户端发生操作时,必须将数据保存在Firebase数据库上,并且必须更新GeoFire 我会用一些代码更清楚 1-客户端的操作。为用户及其位置存储温度Android 移除并重新插入GeoFire位置,以触发onKeyExited和onKeyEntered不';行不通,android,firebase,firebase-realtime-database,geofire,Android,Firebase,Firebase Realtime Database,Geofire,我对GeoFire和Firebase有问题 我想根据GeoFire位置触发器更新UI上的信息。当客户端发生操作时,必须将数据保存在Firebase数据库上,并且必须更新GeoFire 我会用一些代码更清楚 1-客户端的操作。为用户及其位置存储温度 public void sendTemperature(Float temp) { LatLng myPosition = getMyPosition(); //instantiate the db connection F
public void sendTemperature(Float temp) {
LatLng myPosition = getMyPosition();
//instantiate the db connection
FirebaseDatabase database = FirebaseDatabase.getInstance();
DatabaseReference dbTemperatures = database.getReference("temperatures");
//push the value to global temperature db
String tempId = dbTemperatures.push().getKey();
dbTemperatures.child("/"+tempId).setValue(temp);
//create an index to link the user with its last sent temperature
DatabaseReference dbUserLastTemperature = database.getReference("users_last_temperature");
dbUserLastTemperature.child(userId).setValue(tempId);
//update the geofireDb (here is the problem)
geoUsersTemperatures.removeLocation(userId);
geoUsersTemperatures.setLocation(userId, new GeoLocation(myPosition.latitude, myPosition.longitude));
}
2-客户端的侦听器。通知附近用户的温度
GeoQuery usersTemperatureGeoQuery = geoUsersTemperatures.queryAtLocation(new GeoLocation(getMyPosition().latitude, getMyPosition().longitude), mapsRadius);
usersTemperatureGeoQuery.addGeoQueryEventListener(new GeoQueryEventListener() {
@Override
public void onKeyEntered(String key, GeoLocation location) {
System.out.println(String.format("Key %s entered the search area at [%f,%f]", key, location.latitude, location.longitude));
addTemperature(key, location);
}
@Override
public void onKeyExited(String key) {
System.out.println(String.format("Key %s is no longer in the search area", key));
removeTemperature(key);
}
@Override
public void onKeyMoved(String key, GeoLocation location) {
System.out.println(String.format("Key %s moved within the search area to [%f,%f]", key, location.latitude, location.longitude));
moveTemperature(key, location);
}
@Override
public void onGeoQueryReady() {
System.out.println("All initial data has been loaded and events have been fired!");
}
@Override
public void onGeoQueryError(DatabaseError error) {
System.err.println("There was an error with this query: " + error);
}
});
3-问题。我期望发生的是,当一个新的温度被发送时,监听器检测到它的删除和新的插入(但实际上它的键没有改变),所以我可以更新我的UI。这不会发生,因为这两个动作非常快,GeoFire侦听器不会触发任何东西,因此数据库中没有任何更改
有没有办法让侦听器按我所希望的方式运行?这可能可以通过使用所谓的in to
removeLocation()
来解决
未经测试的快速撰写(因此可能包含语法错误):
使用这种方法,在设置新位置之前,您需要等待数据库确认它已删除旧位置
public void sendTemperature(Float temp) {
LatLng myPosition = getMyPosition();
//instantiate the db connection
FirebaseDatabase database = FirebaseDatabase.getInstance();
DatabaseReference dbTemperatures = database.getReference("temperatures");
//push the value to global temperature db
String tempId = dbTemperatures.push().getKey();
dbTemperatures.child("/"+tempId).setValue(temp);
//create an index to link the user with its last sent temperature
DatabaseReference dbUserLastTemperature = database.getReference("users_last_temperature");
dbUserLastTemperature.child(userId).setValue(tempId);
//update the geofireDb (here is the problem)
geoUsersTemperatures.removeLocation(userId);
geoUsersTemperatures.setLocation(userId, new GeoLocation(myPosition.latitude, myPosition.longitude));
}