Question

我的服务需要GPS。我实际上实现了一个服务，它打开gps，检查位置是否有效，然后进入睡眠状态一段时间。（从5秒开始，如果没有检测到移动，它可以睡到一分钟）之后，我再次启动gps，并获得一个新位置。

但是电池耗尽仍然很高！我使用过locMgr.distanceFilter = 10.0f和desiredAccurady = NearestTenMeters。

如何更多地减少电池消耗？

Answer 1

这就是我处理它的方式。在我获取位置后，我将其存储在NSDictionary中。然后如果我需要再次定位我返回NSDictionary而不是重新打开GPS。 2分钟后，我重置了NSDictionary（你可以调整你应用程序的最佳套件的时间）。然后，在NSDictionary重置后我下次需要该位置时，我从GPS获取一个新位置。

- (NSDictionary *) getCurrentLocation {

if (self.currentLocationDict == nil) {
    self.currentLocationDict = [[NSMutableDictionary alloc] init];

    locationManager = [[CLLocationManager alloc] init];
    locationManager.delegate = self;
    locationManager.desiredAccuracy = kCLLocationAccuracyHundredMeters;
    locationManager.distanceFilter = kCLDistanceFilterNone;
    [locationManager startUpdatingLocation];

    CLLocation *myLocation = [locationManager location];

    [self.currentLocationDict setObject:[NSString stringWithFormat:@"%f", myLocation.coordinate.latitude] forKey:@"lat"];
    [self.currentLocationDict setObject:[NSString stringWithFormat:@"%f", myLocation.coordinate.longitude] forKey:@"lng"];
    [locationManager stopUpdatingLocation];
    [locationManager release];

    //Below timer is to help save battery by only getting new location only after 2 min has passed from the last time a new position was taken.  Last location is saved in currentLocationDict
    [NSTimer scheduledTimerWithTimeInterval:120 target:self selector:@selector(resetCurrentLocation) userInfo:nil repeats:NO];
}

return self.currentLocationDict;
}

- (void) resetCurrentLocation {
NSLog(@"reset");
[currentLocationDict release];
self.currentLocationDict = nil;
}

iOS GPS电池耗尽，如何减少耗电量？

1 个答案: