Reputation: 5232

Open application with bundle identifier

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..?

Upvotes: 14

Answers (6)

Dmytro Shvetsov

Reputation: 1006

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

Upvotes: 13

Alok

Reputation: 25978

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

Upvotes: 2

Evan JIANG

Reputation: 710

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"];

Upvotes: 16