How can I open other app from my app without URL scheme if app is exist in my device? [closed]-CodePudding

in iOS swift I have 1000 URL of different apps with bundle Id, I I want to open app if exist in my device, how can I open? I can't get URL scheme if all apps, is there any other apps? Thanks.

CodePudding user response：

Use below function.

func openApp(appId : String) {
    if let url = URL(string: "itms-apps://itunes.apple.com/app/\(appId)"),
        UIApplication.shared.canOpenURL(url)
    {
        if #available(iOS 10.0, *) {
            UIApplication.shared.open(url, options: [:], completionHandler: nil)
        } else {
            UIApplication.shared.openURL(url)
        }
    }
}

Usage

openApp(appId: "id284882215")

This will open facebook app else take to app store.

CodePudding user response：

we can use like this. For example you want to open an Instagram application

let instagram = "instagram://user?username=johndoe"
let instagramUrl = URL(string: instagram)!
if UIApplication.shared.canOpenURL(instagramUrl)
{  
    UIApplication.shared.open(instagramUrl)
} else {
    UIApplication.shared.open(URL(string: "http://instagram.com/")!)
}