Launch an app from within another (iPhone)

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Is it possible to launch any arbitrary iPhone application from within another app?, For example in my application if I want the user to push a button and launch right into the Phone app (close the current app, open the Phone app).

would this be possible? I know this can be done for making phone calls with the tel URL link, but I want to instead just have the Phone app launch without dialing any specific number.

openurl deep-linking iphone cocoa-touch ios
14 Answers

As Kevin points out, URL Schemes are the only way to communicate between apps. So, no, it's not possible to launch arbitrary apps.

But it is possible to launch any app that registers a URL Scheme, whether it's Apple's, yours, or another developer's. The docs are here:

Defining a Custom URL Scheme for Your App

As for launching the phone, looks like your tel: link needs to have least three digits before the phone will launch. So you can't just drop into the app without dialing a number.

Gordon_Wilson_230710

You can only launch apps that have registered a URL scheme. Then just like you open the SMS app by using sms:, you'll be able to open the app using their URL scheme.

There is a very good example available in the docs called LaunchMe which demonstrates this.

LaunchMe sample code as of 6th Nov 2017.

lostInTransit_462970

In Swift 4.1 and Xcode 9.4.1

I have two apps 1)PageViewControllerExample and 2)DelegateExample. Now i want to open DelegateExample app with PageViewControllerExample app. When i click open button in PageViewControllerExample, DelegateExample app will be opened.

For this we need to make some changes in .plist files for both the apps.

Step 1

In DelegateExample app open .plist file and add URL Types and URL Schemes. Here we need to add our required name like "myapp".

enter image description here

Step 2

In PageViewControllerExample app open .plist file and add this code

<key>LSApplicationQueriesSchemes</key>
<array>
    <string>myapp</string>
</array>

Now we can open DelegateExample app when we click button in PageViewControllerExample.

//In PageViewControllerExample create IBAction
@IBAction func openapp(_ sender: UIButton) {

    let customURL = URL(string: "myapp://")
    if UIApplication.shared.canOpenURL(customURL!) {

        //let systemVersion = UIDevice.current.systemVersion//Get OS version
        //if Double(systemVersion)! >= 10.0 {//10 or above versions
            //print(systemVersion)
            //UIApplication.shared.open(customURL!, options: [:], completionHandler: nil)
        //} else {
            //UIApplication.shared.openURL(customURL!)
        //}

        //OR

        if #available(iOS 10.0, *) {
            UIApplication.shared.open(customURL!, options: [:], completionHandler: nil)
        } else {
            UIApplication.shared.openURL(customURL!)
        }
    } else {
         //Print alert here
    }
}
user_80908930

No it's not. Besides the documented URL handlers, there's no way to communicate with/launch another app.

user_5820

A side note regarding this topic...

There is a 50 request limit for protocols that are not registered.

In this discussion apple mention that for a specific version of an app you can only query the canOpenUrl a limited number of times and will fail after 50 calls for undeclared schemes. I've also seen that if the protocol is added once you have entered this failing state it will still fail.

Be aware of this, could be useful to someone.

user_33920170
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