当应用程序使用URL方案在iOS中打开我的应用程序时，如何获取呼叫者的信息

Question

In my situation, a third-party app will using URL Scheme to open my App in iOS, but my App must get some information about the caller, like caller's app name, caller's app icon.

How to do this ? or Is there another way to solve the situation?

Answer 1

If you implement openURL in your AppDelegate , you can get some of those value, app icon is not possible though, you have to prepare it yourself or the scheme must somehow have it as url/data:

func application(application: UIApplication, openURL url: NSURL, sourceApplication: String?, annotation: AnyObject) -> Bool {
        print("source \(sourceApplication)")
        print("scheme \(url.scheme)")
        print("query \(url.query)")
        print("full \(url.absoluteString)")
}

Answer 2

iOS can bind your app and a custom URL Scheme. With this URL Scheme you can
launch your app from browser or other app.

You create the URL Scheme like snapshot:

URL Scheme: myapp
URL identifier: com.ypd.myapp

Then you can send URL like this below:

myapp://
myapp://example/path/here
myapp://?param1=1&param2=2
myapp://some/path/here?param1=1&param2=2

From the delegate method, you can get the deliver data:

- (BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url   
{  
    // Do something with the url here  
}