使用Google地图和Apple Maps URL计划自动开始导航

时间:2014-09-25 13:21:03

标签: ios google-maps url-scheme apple-maps

在iOS上使用URL方案启动Google MapsApple Maps时,是否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。

3 个答案:

答案 0 :(得分:4)

以下是我的参考方式,但是对于苹果,我还没有办法通过网址方案开始导航。

+ (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]];
    }
}

答案 1 :(得分:2)

你是对的,谷歌和苹果都需要用户输入 - 但只需点击“开始”按钮。

如果要同时指定开始和结束位置,请使用以下格式:

Apple Maps:

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

Google地图:

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

答案 2 :(得分:2)

用于启动Apple Maps或Google Maps导航的Swift 3帮助程序类

 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)
    }
}