我将Waze集成到我的Swift应用程序中,但是当我点击按钮时,Waze打开但导航没有任何反应。我会看到应用程序,而不是启动导航。
以下是代码:
@IBAction func openWazeAction(_ sender: Any) {
// open waze
if UIApplication.shared.canOpenURL(URL(string: "waze://")!) {
let urlStr = String(format: "waze://ul?ll=%f,%f&navigate=yes", (selectedBorne?.location?.x)!, (selectedBorne?.location?.y)!)
print(urlStr)
UIApplication.shared.open(URL(string: urlStr)!)
} else {
UIApplication.shared.open(URL(string: "http://itunes.apple.com/us/app/id323229106")!)
}
}
print(urlStr)
会返回正确的网址:waze://ul?ll=48.792914,2.366290&navigate=yes
,但Waze应用中没有任何内容。
(我把LSApplicationQueriesSchemes放在Info.plist文件中。)
这里有什么问题?
答案 0 :(得分:6)
我解决了这个问题。 Waze documentation提供了错误的信息,因为他们的iOS示例无法按原样打开Waze应用。它在移动设备上打开Safari,然后我们需要点击链接打开Waze。
正确的链接是:
waze://?ll={latitude},{longitude}&navigate=yes
我需要删除网址中的ul
。
func navigateTo(latitude: Double, longitude: Double) {
if UIApplication.shared.canOpenURL(URL(string: "waze://")!) {
// Waze is installed. Launch Waze and start navigation
let urlStr = String(format: "waze://?ll=%f,%f&navigate=yes", latitude, longitude)
UIApplication.shared.open(URL(string: urlStr)!)
} else {
// Waze is not installed. Launch AppStore to install Waze app
UIApplication.shared.open(URL(string: "http://itunes.apple.com/us/app/id323229106")!)
}
}
(void) navigateToLatitude:(double)latitude longitude:(double)longitude
{
if ([[UIApplication sharedApplication]
canOpenURL:[NSURL URLWithString:@"waze://"]]) {
// Waze is installed. Launch Waze and start navigation
NSString *urlStr =
[NSString stringWithFormat:@"waze://?ll=%f,%f&navigate=yes",
latitude, longitude];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:urlStr]];
} else {
// Waze is not installed. Launch AppStore to install Waze app
[[UIApplication sharedApplication] openURL:[NSURL
URLWithString:@"http://itunes.apple.com/us/app/id323229106"]];
}
}
答案 1 :(得分:0)
所选答案对我不起作用,我在安装了位智的设备上运行该应用程序,并且始终打开了该应用程序商店。使用此代码,如果已安装,它将打开位智,否则将打开appstore。
let urlStr = String(format: "waze://?ll=%f,%f&navigate=yes", latitude, longitude)
UIApplication.shared.open(URL(string: urlStr)!) { didOpen in
if !didOpen {
UIApplication.shared.open(URL(string: "http://itunes.apple.com/us/app/id323229106")!)
}
}