Open application with bundle identifier

Question

Is it possible to open a application from our application with bundle identifier. Suppose I have two apps installed on device one with com.test.app1 and com.test.app2. Can I open app1 from my app2.

I know about openUrl method. for that I have to register url scheme in info.plist. and then i can use following method:

[[UIApplication sharedApplication] openUrl:[NSURL urlWithString:@"myApp1://"]];

But what if I didn't register url scheme or don't know the registered url.

Any idea..?

Answer 1

I don't think that's possible.

Answer 2

You can use private API to do that

Class LSApplicationWorkspace_class = objc_getClass("LSApplicationWorkspace");
NSObject * workspace = [LSApplicationWorkspace_class performSelector:@selector(defaultWorkspace)];
BOOL isopen = [workspace performSelector:@selector(openApplicationWithBundleID:) withObject:@"com.apple.mobilesafari"];

Answer 3

You can use the openUrl call, but in order to succeed you must add some values to your project's xy-Info.plist file.

Once you've done that you can then call:

[[UIApplication sharedApplication] openUrl:[NSURL urlWithString:@"xingipad://"]];

Answer 4

The Swift version of @EvanJIANG answer.

guard let obj = objc_getClass("LSApplicationWorkspace") as? NSObject else { return false }
let workspace = obj.perform(Selector(("defaultWorkspace")))?.takeUnretainedValue() as? NSObject
let open = workspace?.perform(Selector(("openApplicationWithBundleID:")), with: "com.apple.mobilesafari") != nil
return open

Answer 5

Answer: You can't open app directly only with Bundle identifier.

Solution: You can implement deep linking (and take your bundle id as your deep linking id)concept to do this: Deep-linking

Answer 6

It is possible using URL Schemes .