在swift中撥打電話號碼

Question

我正在嘗試撥打一個不使用特定號碼的號碼，而是撥打一個在變量中被呼叫的號碼，或者至少告訴它在您的手機中撥打號碼。這個在變量中被調用的數字是我通過使用解析器或從網站sql獲取的數字。我做了一個按鈕，試圖用函數調用存儲在變量中的電話號碼，但無濟於事。任何事情都會有幫助謝謝！

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)!) 

     } 

Answer 1

試試看：

if let url = NSURL(string: "tel://\(busPhone)") where UIApplication.sharedApplication().canOpenURL(url) { 
    UIApplication.sharedApplication().openURL(url) 
} 

假設的電話號碼是busPhone。

NSURL的init(string:)返回一個可選的，所以通過使用if let我們確保url是NSURL（而不是NSURL?由init返回）。

---

對於斯威夫特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) 
    } 
} 

我們需要檢查我們是否是iOS上的10或更高版本，因爲：

'openURL' was deprecated in iOS 10.0

Answer 2

好我得到的幫助和理解了它。此外，爲了防止電話號碼無效，我還提供了一個很好的小警報系統。我的問題是我打電話給它的權利，但數字有空格和不需要的字符，如（「123 456-7890」）。如果您的號碼是（「1234567890」），則UIApplication僅適用或接受。所以你基本上通過創建一個新的變量來只取出數字來刪除空格和無效字符。然後用UIApplication調用這些數字。

func callSellerPressed (sender: UIButton!){ 
     var newPhone = "" 

     for (var i = 0; i < 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 > 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() 
     } 
     } 

Answer 3

在iOS的10自載的解決方案，斯威夫特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) 
    } 
    } 
} 

你應該能夠使用callNumber("7178881234")撥打電話。

Answer 4

這是@Tom的回答使用Swift 2.0的更新 注意 - 這是我使用的整個CallComposer類。

class CallComposer: NSObject { 

var editedPhoneNumber = "" 

func call(phoneNumber: String) -> 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 
} 
} 

Answer 5

我在我的應用程序中使用這種方法，它工作正常。我希望這可以幫助你。

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) 
} 

Answer 6

夫特3.0溶液：

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) 

Answer 7

在夫特3，

if let url = URL(string:"tel://\(phoneNumber)"), UIApplication.shared.canOpenURL(url) { 
    UIApplication.shared.openURL(url) 
} 

Answer 8

上述答案是部分正確，但與 「電話：//」 只存在一個問題。通話結束後，它將返回到主屏幕，而不是我們的應用程序。所以最好使用「telprompt：//」，它會返回到應用程序。

var url:NSURL = NSURL(string: "telprompt://1234567891")! 
UIApplication.sharedApplication().openURL(url) 

Answer 9

的OpenURL（）已在IOS 10.這裏被棄用是新的語法：

if let url = URL(string: "tel://\(busPhone)") { 
    UIApplication.shared.open(url, options: [:], completionHandler: nil) 
} 

Answer 10

夫特3.0和IOS 10或以上

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) 
     } 
    } 
} 

Answer 11

這裏的減少的另一種方式使用Scanner的有效元件的電話號碼...

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: &digits) 
    } else { 
     scanner.scanCharacters(from: validCharacters, into: &digits) 
    } 

    scanner.scanUpToCharacters(from: validCharacters, into: nil) 
    if let digits = digits as? String { 
     validNumber.append(digits) 
    } 
} 

print(validNumber) 

// +1234567890 

Answer 12

我使用SWIFT 3溶液用數字驗證

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)")!) 
    } 

Answer 13

雨燕3.0 &的iOS 10+

UIApplication.shared.openURL(url) 改爲 UIApplication.shared.open(_ url: URL, options:[:], completionHandler completion: nil)

選項和完成處理是可選的，呈現：

UIApplication.shared.open(url)

https://developer.apple.com/reference/uikit/uiapplication/1648685-open

Answer 14

夫特3，iOS裝置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) 
      } 
     } 
    } 

Answer 15

對於夫特3.1 &向後兼容的方法，這樣做：

@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) 
    } 
} 

Answer 16

的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.") 
}