Launch an app from within another iPhone

Question

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

User · Answer

Try the following code will help you to Launch an application from your application

Note: Replace the name fantacy with actual application name

NSString *mystr=[[NSString alloc] initWithFormat:@"fantacy://location?id=1"];
NSURL *myurl=[[NSURL alloc] initWithString:mystr];
[[UIApplication sharedApplication] openURL:myurl];

User · Answer

In order to let you open your application from another  you ll need to make changes in both applications  Here are the steps using Swift 3 with iOS 10 update   1  Register your application that you want to open  Update the Info plist by defining your application s custom and unique URL Scheme     Note that your scheme name should be unique  otherwise if you have another application with the same URL scheme name installed on your device  then this will be determined runtime which one gets opened   2  Include the previous URL scheme in your main application  You ll need to specify the URL scheme you want the app to be able to use with the canOpenURL  method of the UIApplication class  So open the main application s Info plist and add the other application s URL scheme to LSApplicationQueriesSchemes   Introduced in iOS 9 0     3  Implement the action that opens your application  Now everything is set up  so you re good to write your code in your main application that opens your other app  This should looks something like this   let appURLScheme    MyAppToOpen      guard let appURL   URL string  appURLScheme  else       return    if UIApplication shared canOpenURL appURL         if  available iOS 10 0               UIApplication shared open appURL            else           UIApplication shared openURL appURL          else          Here you can handle the case when your other application cannot be opened for any reason      Note that these changes requires a new release if you want your existing app  installed from AppStore  to open  If you want to open an application that you ve already released to Apple AppStore  then you ll need to upload a new version first that includes your URL scheme registration

User · Answer

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 · Answer

In order to let you open your application from another  you ll need to make changes in both applications  Here are the steps using Swift 3 with iOS 10 update   1  Register your application that you want to open  Update the Info plist by defining your application s custom and unique URL Scheme     Note that your scheme name should be unique  otherwise if you have another application with the same URL scheme name installed on your device  then this will be determined runtime which one gets opened   2  Include the previous URL scheme in your main application  You ll need to specify the URL scheme you want the app to be able to use with the canOpenURL  method of the UIApplication class  So open the main application s Info plist and add the other application s URL scheme to LSApplicationQueriesSchemes   Introduced in iOS 9 0     3  Implement the action that opens your application  Now everything is set up  so you re good to write your code in your main application that opens your other app  This should looks something like this   let appURLScheme    MyAppToOpen      guard let appURL   URL string  appURLScheme  else       return    if UIApplication shared canOpenURL appURL         if  available iOS 10 0               UIApplication shared open appURL            else           UIApplication shared openURL appURL          else          Here you can handle the case when your other application cannot be opened for any reason      Note that these changes requires a new release if you want your existing app  installed from AppStore  to open  If you want to open an application that you ve already released to Apple AppStore  then you ll need to upload a new version first that includes your URL scheme registration

User · Answer

To achieve this we need to add few line of Code in both App  App A  Which you want to open from another App   Source   App B  From App B you want to open App A  Destination   Code for App A  Add few tags into the Plist of App A Open Plist Source of App A and Past below XML   lt key gt CFBundleURLTypes lt  key gt   lt array gt       lt dict gt           lt key gt CFBundleURLName lt  key gt           lt string gt com TestApp lt  string gt           lt key gt CFBundleURLSchemes lt  key gt           lt array gt               lt string gt testApp linking lt  string gt           lt  array gt       lt  dict gt   lt  array gt    In App delegate of App A - Get Callback here  -  BOOL application  UIApplication   application openURL  NSURL   url   sourceApplication  NSString   sourceApplication annotation  id annotation      You we get the call back here when App B will try to Open     sourceApplication will have the bundle ID of the App B     url query  will provide you the whole URL      url query  with the help of this you can also pass the value from App B and get that value here      Now coming to App B code -   If you just want to open App A without any input parameter   - IBAction openApp A  id sender       if    UIApplication sharedApplication  openURL  NSURL URLWithString   testApp linking                   UIAlertView  alert     UIAlertView alloc initWithTitle   App is not available   message nil delegate self cancelButtonTitle   Ok  otherButtonTitles nil  nil                alert show                  If you want to pass parameter from App B to App A then use below Code  - IBAction openApp A  id sender      if    UIApplication sharedApplication  openURL  NSURL URLWithString   testApp linking    userName abe amp registered 1 amp Password 123abc               UIAlertView  alert     UIAlertView alloc initWithTitle   App is not available   message nil delegate self cancelButtonTitle   Ok  otherButtonTitles nil  nil                alert show                  Note  You can also open App with just type testApp linking     on safari browser

User · Answer

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

User · Answer

I also tried this a while ago  Launch iPhone Application with Identifier   but there definitely is no DOCUMENTED way to do this

User · Answer

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

User · Answer

Here is a good tutorial for launching application from within another app  iOS SDK  Working with URL Schemes And  it is not possible to launch arbitrary application  but the native applications which registered the URL Schemes

User · Answer

I found that it s easy to write an app that can open another app    Let s assume that we have two apps called FirstApp and SecondApp  When we open the FirstApp  we want to be able to open the SecondApp by clicking a button  The solution to do this is     In SecondApp  Go to the plist file of SecondApp and you need to add a URL Schemes with a string iOSDevTips of course you can write another string it s up to you       2   In FirstApp  Create a button with the below action   -  void buttonPressed  UIButton   button     NSString  customURL     iOSDevTips         if    UIApplication sharedApplication  canOpenURL  NSURL URLWithString customURL              UIApplication sharedApplication  openURL  NSURL URLWithString customURL          else         UIAlertView  alert     UIAlertView alloc  initWithTitle   URL error                                message  NSString stringWithFormat   No custom URL defined for      customURL                                delegate self cancelButtonTitle   Ok                                 otherButtonTitles nil        alert show            That s it  Now when you can click the button in the FirstApp it should open the SecondApp

User · Answer

In Swift  Just in case someone was looking for a quick Swift copy and paste   if let url   NSURL string   app      where UIApplication sharedApplication   canOpenURL url                UIApplication sharedApplication   openURL url    else if let itunesUrl   NSURL string   https   itunes apple com itunes-link-to-app   where UIApplication sharedApplication   canOpenURL itunesUrl                UIApplication sharedApplication   openURL itunesUrl

User · Answer

I also tried this a while ago  Launch iPhone Application with Identifier   but there definitely is no DOCUMENTED way to do this

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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      Step 2  In PageViewControllerExample app open  plist file and add this code   lt key gt LSApplicationQueriesSchemes lt  key gt   lt array gt       lt string gt myapp lt  string gt   lt  array gt    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    gt   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 · Answer

In Swift  Just in case someone was looking for a quick Swift copy and paste   if let url   NSURL string   app      where UIApplication sharedApplication   canOpenURL url                UIApplication sharedApplication   openURL url    else if let itunesUrl   NSURL string   https   itunes apple com itunes-link-to-app   where UIApplication sharedApplication   canOpenURL itunesUrl                UIApplication sharedApplication   openURL itunesUrl

User · Answer

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

User · Answer

Try the following code will help you to Launch an application from your application

Note: Replace the name fantacy with actual application name

NSString *mystr=[[NSString alloc] initWithFormat:@"fantacy://location?id=1"];
NSURL *myurl=[[NSURL alloc] initWithString:mystr];
[[UIApplication sharedApplication] openURL:myurl];

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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      Step 2  In PageViewControllerExample app open  plist file and add this code   lt key gt LSApplicationQueriesSchemes lt  key gt   lt array gt       lt string gt myapp lt  string gt   lt  array gt    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    gt   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 · Answer

Here is a good tutorial for launching application from within another app  iOS SDK  Working with URL Schemes And  it is not possible to launch arbitrary application  but the native applications which registered the URL Schemes

User · Answer

I found that it s easy to write an app that can open another app    Let s assume that we have two apps called FirstApp and SecondApp  When we open the FirstApp  we want to be able to open the SecondApp by clicking a button  The solution to do this is     In SecondApp  Go to the plist file of SecondApp and you need to add a URL Schemes with a string iOSDevTips of course you can write another string it s up to you       2   In FirstApp  Create a button with the below action   -  void buttonPressed  UIButton   button     NSString  customURL     iOSDevTips         if    UIApplication sharedApplication  canOpenURL  NSURL URLWithString customURL              UIApplication sharedApplication  openURL  NSURL URLWithString customURL          else         UIAlertView  alert     UIAlertView alloc  initWithTitle   URL error                                message  NSString stringWithFormat   No custom URL defined for      customURL                                delegate self cancelButtonTitle   Ok                                 otherButtonTitles nil        alert show            That s it  Now when you can click the button in the FirstApp it should open the SecondApp

User · Answer

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

User · Answer

Swift 3 quick and paste version of  villy393 answer for      if UIApplication shared canOpenURL openURL        UIApplication shared openURL openURL    else if UIApplication shared canOpenURL installUrl       UIApplication shared openURL installUrl

User · Answer

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 · Answer

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

User · Answer

Swift 3 quick and paste version of  villy393 answer for      if UIApplication shared canOpenURL openURL        UIApplication shared openURL openURL    else if UIApplication shared canOpenURL installUrl       UIApplication shared openURL installUrl

User · Answer

The lee answer is absolutely correct for iOS prior to 8   In iOS 9  you must whitelist any URL schemes your App wants to query in Info plist under the LSApplicationQueriesSchemes key  an array of strings

User · Answer

To achieve this we need to add few line of Code in both App  App A  Which you want to open from another App   Source   App B  From App B you want to open App A  Destination   Code for App A  Add few tags into the Plist of App A Open Plist Source of App A and Past below XML   lt key gt CFBundleURLTypes lt  key gt   lt array gt       lt dict gt           lt key gt CFBundleURLName lt  key gt           lt string gt com TestApp lt  string gt           lt key gt CFBundleURLSchemes lt  key gt           lt array gt               lt string gt testApp linking lt  string gt           lt  array gt       lt  dict gt   lt  array gt    In App delegate of App A - Get Callback here  -  BOOL application  UIApplication   application openURL  NSURL   url   sourceApplication  NSString   sourceApplication annotation  id annotation      You we get the call back here when App B will try to Open     sourceApplication will have the bundle ID of the App B     url query  will provide you the whole URL      url query  with the help of this you can also pass the value from App B and get that value here      Now coming to App B code -   If you just want to open App A without any input parameter   - IBAction openApp A  id sender       if    UIApplication sharedApplication  openURL  NSURL URLWithString   testApp linking                   UIAlertView  alert     UIAlertView alloc initWithTitle   App is not available   message nil delegate self cancelButtonTitle   Ok  otherButtonTitles nil  nil                alert show                  If you want to pass parameter from App B to App A then use below Code  - IBAction openApp A  id sender      if    UIApplication sharedApplication  openURL  NSURL URLWithString   testApp linking    userName abe amp registered 1 amp Password 123abc               UIAlertView  alert     UIAlertView alloc initWithTitle   App is not available   message nil delegate self cancelButtonTitle   Ok  otherButtonTitles nil  nil                alert show                  Note  You can also open App with just type testApp linking     on safari browser

User · Answer

The lee answer is absolutely correct for iOS prior to 8   In iOS 9  you must whitelist any URL schemes your App wants to query in Info plist under the LSApplicationQueriesSchemes key  an array of strings

[ios] Launch an app from within another (iPhone)

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