使用Google地图和Apple Maps URL计划自动开始导航

Question

在iOS上使用URL方案启动Google地图或Apple地图时，有没有办法自动启动导航？

我看到两个可选参数，但没有用户输入就没有开始导航。

Answer 1

这是我如何为你的参考，但对于苹果，我还没有找到一种方法来开始通过网址方案导航。

+ (void)navigateToLocation:(CLLocation*)_navLocation {
    if ([[UIApplication sharedApplication] canOpenURL:
         [NSURL URLWithString:@"comgooglemaps://"]]) {
        NSString *string = [NSString stringWithFormat:@"comgooglemaps://?daddr=%f,%f&directionsmode=driving",_navLocation.coordinate.latitude,_navLocation.coordinate.longitude];
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:string]];
    } else {
        NSString *string = [NSString stringWithFormat:@"http://maps.apple.com/?ll=%f,%f",_navLocation.coordinate.latitude,_navLocation.coordinate.longitude];
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:string]];
    }
}

Answer 2

你是对的，谷歌和苹果都需要用户输入 - 但只能点击go按钮。

如果要同时指定起始位置和结束位置，请使用以下格式：

Apple地图：

http://maps.apple.com/maps?saddr=Current%20Location&daddr=<Your Location>

谷歌地图：

comgooglemaps-x-callback://?saddr=&daddr=<Your Location>

Answer 3

用于启动Apple Maps或Google Maps导航的Swift 3辅助类

 struct LinksHelper {
    static func startNavigation(coordinate: CLLocationCoordinate2D) {
        struct Links {
            static let kGoogleMapsSchema = "comgooglemaps://"
            static let kGoogleMaps = "\(kGoogleMapsSchema)?daddr=%f,%f&directionsmode=driving"
            static let kAppleMaps = "https://maps.apple.com/?saddr=Current Location&daddr=%f,%f&z=10&t=s"
        }

        var path: String!

        if let googleMapsSchemaUrl = URL(string:Links.kGoogleMapsSchema), UIApplication.shared.canOpenURL(googleMapsSchemaUrl) {
            path = Links.kGoogleMaps
        } else {
            path = Links.kAppleMaps
        }
        guard let str = String(format: path, coordinate.latitude, coordinate.longitude).addingPercentEncoding(
            withAllowedCharacters: .urlQueryAllowed) else {
                return
        }

        guard let url = URL(string: str) else {
            return
        }

        UIApplication.shared.openURL(url)
    }
}