Calling a phone number in swift

Question

I m trying to call a number not using specific numbers but a number that is being called in a variable or at least tell it to pull up the number in your phone  This number that is being called in a variable is a number that I retrieved by using a parser or grabbing from a website sql  I made a button trying to call the phone number stored in the variable with a function but to no avail  Anything will help thanks       func callSellerPressed  sender  UIButton        This is calls a specific number UIApplication sharedApplication   openURL NSURL string   tel                   This is the code I m using but its not working        UIApplication sharedApplication   openURL NSURL scheme  NSString    host   tel      path  busPhone

User · Answer

For swift 3 0  if let url   URL string   tel     number     UIApplication shared canOpenURL url        if  available iOS 10               UIApplication shared open url        else           UIApplication shared openURL url          else       print  Your device doesn t support this feature

User · Answer

Okay I got help and figured it out  Also I put in a nice little alert system just in case the phone number is not valid  My issue was I was calling it right but the number had spaces and unwanted characters such as   123 456-7890    UIApplication only works or accepts if your number is   1234567890    So you basically remove the space and invalid characters by making a new variable to pull only the numbers  Then calls those numbers with the UIApplication   func callSellerPressed  sender  UIButton            var newPhone               for  var i   0  i  lt  countElements busPhone   i                  var current Int   i             switch  busPhone i                    case  0   1   2   3   4   5   6   7   8   9    newPhone   newPhone   String busPhone i                   default   println  Removed invalid character                                     if   busPhone utf16Count  gt  1            UIApplication sharedApplication   openURL NSURL string   tel       newPhone                      else              let alert   UIAlertView               alert title    Sorry               alert message    Phone number is not available for this business              alert addButtonWithTitle  Ok                   alert show

User · Answer

A self contained solution in iOS 10  Swift 3     private func callNumber phoneNumber String       if let phoneCallURL   URL string   tel     phoneNumber           let application UIApplication   UIApplication shared     if  application canOpenURL phoneCallURL             application open phoneCallURL  options       completionHandler  nil                You should be able to use callNumber  7178881234   to make a call

User · Answer

Just try   if let url   NSURL string   tel     busPhone    where UIApplication sharedApplication   canOpenURL url      UIApplication sharedApplication   openURL url      assuming that the phone number is in busPhone   NSURL s init string   returns an Optional  so by using if let we make sure that url is a NSURL  and not a NSURL  as returned by the init      For Swift 3   if let url   URL string   tel     busPhone     UIApplication shared canOpenURL url        if  available iOS 10               UIApplication shared open url        else           UIApplication shared openURL url            We need to check whether we re on iOS 10 or later because       openURL  was deprecated in iOS 10 0

User · Answer

If your phone number contains spaces  remove them first  Then you can use the accepted answer s solution   let numbersOnly   busPhone replacingOccurrences of       with       if let url   URL string   tel     numbersOnly     UIApplication shared canOpenURL url        if  available iOS 10               UIApplication shared open url        else           UIApplication shared openURL url

User · Answer

The above answers are partially correct  but with  tel     there is only one issue  After the call has ended  it will return to the homescreen  not to our app  So better to use  telprompt      it will return to the app   var url NSURL   NSURL string   telprompt   1234567891    UIApplication sharedApplication   openURL url

User · Answer

I am using swift 3 solution with number validation  var validPhoneNumber          phoneNumber characters forEach   character  in         switch character           case  0     9               validPhoneNumber characters append character          default              break                      if UIApplication shared canOpenURL URL string   tel     validNumber               UIApplication shared openURL URL string   tel     validNumber

User · Answer

This is an update to  Tom s answer using Swift 2 0  Note - This is the whole CallComposer class I am using   class CallComposer  NSObject    var editedPhoneNumber       func call phoneNumber  String  - gt  Bool        if phoneNumber                  for i in number characters                switch  i                   case  0   1   2   3   4   5   6   7   8   9    editedPhoneNumber   editedPhoneNumber   String i                  default   print  Removed invalid character                                 let phone    tel       editedPhoneNumber         let url   NSURL string  phone          if let url   url               UIApplication sharedApplication   openURL url            else               print  There was an error                   else           return false            return true

User · Answer

Swift 3  iOS 10  func call phoneNumber String            let cleanPhoneNumber   phoneNumber components separatedBy  CharacterSet decimalDigits inverted  joined separator              let urlString String    tel     cleanPhoneNumber           if let phoneCallURL   URL string  urlString                if  UIApplication shared canOpenURL phoneCallURL                     UIApplication shared open phoneCallURL  options       completionHandler  nil

User · Answer

Swift 3 0 and ios 10 or older   func phone phoneNum  String        if let url   URL string   tel     phoneNum              if  available iOS 10                   UIApplication shared open url  options       completionHandler  nil            else               UIApplication shared openURL url as URL

User · Answer

let formatedNumber   phone components separatedBy  NSCharacterSet decimalDigits inverted  joined separator      print  calling   formatedNumber    let phoneUrl    tel     formatedNumber   let url URL   URL string  phoneUrl   UIApplication shared openURL url

User · Answer

For a Swift 3 1  amp  backwards compatible approach  do this     IBAction func phoneNumberButtonTouched   sender  Any      if let number   place  phoneNumber       makeCall phoneNumber  number         func makeCall phoneNumber  String       let formattedNumber   phoneNumber components separatedBy      NSCharacterSet decimalDigits inverted  joined separator          let phoneUrl    tel     formattedNumber      let url NSURL   NSURL string  phoneUrl       if  available iOS 10             UIApplication shared open url as URL  options       completionHandler         nil       else        UIApplication shared openURL url as URL

User · Answer

Swift 3 0 solution   let formatedNumber   phone components separatedBy  NSCharacterSet decimalDigits inverted  joined separator      print  calling   formatedNumber    let phoneUrl    tel     formatedNumber   let url URL   URL string  phoneUrl   UIApplication shared openURL url

User · Answer

For Swift 4 2 and above   if let phoneCallURL   URL string   tel     01234567     UIApplication shared canOpenURL phoneCallURL        UIApplication shared open phoneCallURL  options       completionHandler  nil

User · Answer

openURL   has been deprecated in iOS 10  Here is the new syntax   if let url   URL string   tel     busPhone          UIApplication shared open url  options       completionHandler  nil

User · Answer

In Swift 3   if let url   URL string  tel     phoneNumber     UIApplication shared canOpenURL url         UIApplication shared openURL url

User · Answer

Here s an alternative way to reduce a phone number to valid components using a Scanner     let number     123 456-7890   let scanner   Scanner string  number   let validCharacters   CharacterSet decimalDigits let startCharacters   validCharacters union CharacterSet charactersIn          var digits  NSString  var validNumber      while  scanner isAtEnd       if scanner scanLocation    0           scanner scanCharacters from  startCharacters  into   amp digits        else           scanner scanCharacters from  validCharacters  into   amp digits             scanner scanUpToCharacters from  validCharacters  into  nil      if let digits   digits as  String           validNumber append digits           print validNumber       1234567890

User · Answer

Swift 5  iOS  gt   10 0 Only works on physical device  private func callNumber phoneNumber  String        guard let url   URL string   quot telprompt     phoneNumber  quot            UIApplication shared canOpenURL url  else           return           UIApplication shared open url  options       completionHandler  nil

User · Answer

Many of the other answers don t work for Swift 5  Below is the code update to Swift 5   let formattedNumber   phoneNumberVariable components separatedBy  NSCharacterSet decimalDigits inverted  joined separator       if let url   NSURL string    tel      formattedNumber           if  available iOS 10 0               UIApplication shared open url as URL  options       completionHandler  nil        else           UIApplication shared openURL url as URL            PS     With most of the answers  I wasn t able to get the prompt on the device  The above code was successfully able to display the prompt  There is no    after tel  like most answers have  And it works fine

User · Answer

Swift 4    private func callNumber phoneNumber String         if let phoneCallURL   URL string   telprompt     phoneNumber               let application UIApplication   UIApplication shared         if  application canOpenURL phoneCallURL                 if  available iOS 10 0                       application open phoneCallURL  options       completionHandler  nil                else                      Fallback on earlier versions                  application openURL phoneCallURL as URL

User · Answer

I am using this method in my application and it s working fine  I hope this may help you too   func makeCall phone  String        let formatedNumber   phone componentsSeparatedByCharactersInSet NSCharacterSet decimalDigitCharacterSet   invertedSet  joinWithSeparator         let phoneUrl    tel     formatedNumber       let url NSURL   NSURL string  phoneUrl       UIApplication sharedApplication   openURL url

User · Answer

Swift 3 0  amp  iOS 10   UIApplication shared openURL url  was changed to UIApplication shared open   url  URL  options      completionHandler completion  nil   options and completion handler are optional  rendering   UIApplication shared open url   https   developer apple com reference uikit uiapplication 1648685-open

[ios] Calling a phone number in swift

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