我的应用在iOS 7上运行良好,但在iOS 8 SDK上却无法运行。

CLLocationManager不返回位置,我没有看到我的应用程序下设置->位置服务。我在这个问题上做了谷歌搜索,但没有任何结果。会有什么问题呢?


当前回答

        // ** Don't forget to add NSLocationWhenInUseUsageDescription in MyApp-Info.plist and give it a string

        self.locationManager = [[CLLocationManager alloc] init];
        self.locationManager.delegate = self;
        // Check for iOS 8. Without this guard the code will crash with "unknown selector" on iOS 7.
        if ([self.locationManager respondsToSelector:@selector(requestWhenInUseAuthorization)]) {
            [self.locationManager requestWhenInUseAuthorization];
        }
        [self.locationManager startUpdatingLocation];


    // Location Manager Delegate Methods    
    - (void)locationManager:(CLLocationManager *)manager didUpdateLocations:(NSArray *)locations
    {
        NSLog(@"%@", [locations lastObject]);

}

其他回答

解决方案与向后兼容,不产生Xcode警告:

SEL requestSelector = NSSelectorFromString(@"requestWhenInUseAuthorization");
if ([CLLocationManager authorizationStatus] == kCLAuthorizationStatusNotDetermined &&
  [self.locationManager respondsToSelector:requestSelector]) {
((void (*)(id, SEL))[self.locationManager methodForSelector:requestSelector])(self.locationManager, requestSelector);
  [self.locationManager startUpdatingLocation];
} else {
  [self.locationManager startUpdatingLocation];
}

在Info.plist中设置NSLocationWhenInUseUsageDescription键。

对于iOS版本11.0+: 在Info.plist中设置NSLocationAlwaysAndWhenInUseUsageDescription键。还有另外2把钥匙。

我最终解决了自己的问题。

显然,在iOS 8 SDK中,在开始位置更新之前需要调用CLLocationManager上的requestAlwaysAuthorization(用于后台位置)或requestWhenInUseAuthorization(仅用于前台位置)。

在Info中还需要NSLocationAlwaysUsageDescription或NSLocationWhenInUseUsageDescription键。Plist并在提示符中显示一条消息。加上这些就解决了我的问题。

更多详细信息,请查看:Core-Location-Manager-Changes-in-ios-8

用于询问位置的旧代码将无法在iOS 8中使用。您可以尝试使用此方法进行位置授权:

- (void)requestAlwaysAuthorization
{
    CLAuthorizationStatus status = [CLLocationManager authorizationStatus];

    // If the status is denied or only granted for when in use, display an alert
    if (status == kCLAuthorizationStatusAuthorizedWhenInUse || status ==        kCLAuthorizationStatusDenied) {
        NSString *title;
        title = (status == kCLAuthorizationStatusDenied) ? @"Location services are off" :   @"Background location is not enabled";
        NSString *message = @"To use background location you must turn on 'Always' in the Location Services Settings";

        UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:title
                                                            message:message
                                                           delegate:self
                                                  cancelButtonTitle:@"Cancel"
                                                  otherButtonTitles:@"Settings", nil];
        [alertView show];
    }
    // The user has not enabled any location services. Request background authorization.
    else if (status == kCLAuthorizationStatusNotDetermined) {
        [self.locationManager requestAlwaysAuthorization];
    }
}

- (void)alertView:(UIAlertView *)alertView clickedButtonAtIndex:(NSInteger)buttonIndex
{
    if (buttonIndex == 1) {
        // Send the user to the Settings for this app
        NSURL *settingsURL = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
        [[UIApplication sharedApplication] openURL:settingsURL];
    }
}

我也因为同样的问题而抓狂。Xcode给出错误:

试图启动MapKit位置更新而没有提示 位置授权。必须调用-[CLLocationManager requestWhenInUseAuthorization]或-[CLLocationManager . request requestAlwaysAuthorization]。

但即使实现了上述方法之一,它也不会提示用户,除非信息中有条目。NSLocationAlwaysUsageDescription或NSLocationWhenInUseUsageDescription。

添加以下行到您的信息。Plist,其中字符串值代表您需要访问用户位置的原因

<key>NSLocationWhenInUseUsageDescription</key>
<string>This application requires location services to work</string>

<key>NSLocationAlwaysUsageDescription</key>
<string>This application requires location services to work</string>

我认为自从我在Xcode 5中开始这个项目以来,这些条目可能已经丢失了。我猜Xcode 6可能会为这些键添加默认条目,但还没有确认。

你可以在这里找到更多关于这两个设置的信息

对我来说,问题是CLLocationManagerDelegate类是私有的,这阻止了所有的委托方法被调用。我想这不是很常见的情况,但我想我应该提到它,以防它能帮助到任何人。