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..?

score 12 · Answer 1 · answered Jun 01 '16 at 01:19

12

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"];

answered Jun 01 '16 at 01:19

Evan JIANG

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Thank you very much! Works on iOS 11, and don't need to know URL scheme. – DmitryKanunnikoff Jul 26 '17 at 10:35
Except you should never use private APIs. – ekscrypto Oct 17 '17 at 03:41
@ekscrypto But I think it's the only way to match the requirements. And by the way, we successfully uploaded app with this code to AppStore. – Evan JIANG Oct 17 '17 at 10:01
being able to snoop a private api usage by the app review team should not be seen as an authorization to use a private api. – ekscrypto Oct 17 '17 at 10:56
I get the error ```Implicitly declaring library function 'objc_getClass' with type 'id (const char *)'``` on the first line – Luís Cunha Jul 13 '18 at 10:23
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@LuísCunha `#import ` on the top of the file – MEnnabah Sep 18 '19 at 09:29

score 5 · Answer 2 · answered Jul 28 '20 at 20:11

5

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

answered Jul 28 '20 at 20:11

Dmytro Shvetsov

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this worked for me, will my app gonna pass Apple review with this? – Zeeshan Ahmed Jun 16 '21 at 00:44
I am using this just to open different apps using their bundle id's – Zeeshan Ahmed Jun 16 '21 at 00:45
this works. Just tested today. running latest version of Xcode and on device I have running iOS 15.3.1 (19D52) – zumzum Mar 23 '22 at 22:54

leviathan · Answer 3 · 2014-05-20T05:33:38.477

4

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

enter image description here

Once you've done that you can then call:

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

edited May 20 '14 at 05:33

answered Feb 07 '13 at 20:35

leviathan

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score 3 · Accepted Answer · edited Feb 27 '13 at 15:17

3

I don't think that's possible.

edited Feb 27 '13 at 15:17

Tim Cooper

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answered Aug 29 '11 at 12:05

Akshay

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Are you sure that this is not possible..? Because we can check if application is installed in device or not with bundle identifier. so I was wondering that there may be some method for open it. – Kapil Choubisa Aug 29 '11 at 13:17
I don't think an Apple approved way exists. – Akshay Aug 29 '11 at 14:33

Alok · Answer 5 · 2018-06-02T08:55:02.100

0

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

edited Jun 02 '18 at 08:55

answered May 24 '18 at 04:47

Alok

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The question already included deep linking, asker was trying to find a way around that – ekscrypto Jul 18 '18 at 13:20
All steps of deep linking are given in the above answer. – Alok Jul 23 '18 at 02:51

score -1 · Answer 6 · answered Aug 06 '12 at 10:15

-1

It is possible using URL Schemes .

answered Aug 06 '12 at 10:15

krishnendra

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I know that this is possible with URL Scheme but I want to know that is this possible using bundle identifier?? If I don't have url scheme register than is this possible to open an app. – Kapil Choubisa Aug 07 '12 at 06:58

Open application with bundle identifier

6 Answers6