如何使用Swift在iOS 10中拨打电话？

Question

我希望我的应用程序能够在单击按钮时调用某个数字.我试图谷歌它但是到目前为止似乎没有iOS 10(openURL已经消失).有人能为我举个例子吗？例如:

@IBAction func callPoliceButton(_ sender: UIButton) {
    // Call the local Police department
}

Answer 1

你可以这样打电话:

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

对于Swift 3+,您可以使用

guard let number = URL(string: "tel://" + number) else { return }
UIApplication.shared.open(number)

OR

UIApplication.shared.open(number, options: [:], completionHandler: nil)

请确保你已经擦洗你的手机串号删除的任何实例(,),-,或space.

Answer 2

任务

通过电话号码验证拨打电话

细节

Xcode 9.2,Swift 4

解

extension String {

    enum RegularExpressions: String {
        case phone = "^\\s*(?:\\+?(\\d{1,3}))?([-. (]*(\\d{3})[-. )]*)?((\\d{3})[-. ]*(\\d{2,4})(?:[-.x ]*(\\d+))?)\\s*$"
    }

    func isValid(regex: RegularExpressions) -> Bool { return isValid(regex: regex.rawValue) }
    func isValid(regex: String) -> Bool { return range(of: regex, options: .regularExpression) != nil }

    func onlyDigits() -> String {
        let filtredUnicodeScalars = unicodeScalars.filter { CharacterSet.decimalDigits.contains($0) }
        return String(String.UnicodeScalarView(filtredUnicodeScalars))
    }

    func makeAColl() {
        guard   isValid(regex: .phone),
                let url = URL(string: "tel://\(self.onlyDigits())"),
                UIApplication.shared.canOpenURL(url) else { return }
        if #available(iOS 10, *) {
            UIApplication.shared.open(url)
        } else {
            UIApplication.shared.openURL(url)
        }
    }
}

用法

"+1-(800)-123-4567".makeAColl()

样品进行测试

"phone:+1(617)111-22-33!".makeAColl() // Will extract "+1(617)111-22-33" and make a call

结果

Answer 3

针对Swift 3进行了更新:

如果您想拨打电话,请使用以下简单的代码行:

//功能定义:

func makeAPhoneCall()  {
    let url: NSURL = URL(string: "TEL://1234567890")! as NSURL
    UIApplication.shared.open(url as URL, options: [:], completionHandler: nil)
}

//函数调用:[在代码中的任何位置使用]

self.makeAPhoneCall()

注意:请在真实设备上运行该应用,因为它无法在模拟器上运行.



Answer 4

在Swift 4.2中

func dialNumber(number : String) {

 if let url = URL(string: "tel://\(number)"),
   UIApplication.shared.canOpenURL(url) {
      if #available(iOS 10, *) {
        UIApplication.shared.open(url, options: [:], completionHandler:nil)
       } else {
           UIApplication.shared.openURL(url)
       }
   } else {
            // add error message here 
   }
}

如下所示

dialNumber(number: "+921111111222")

希望这有帮助.

Answer 5

if let phoneCallURL:URL = URL(string: "tel:\(strPhoneNumber)") {
        let application:UIApplication = UIApplication.shared
        if (application.canOpenURL(phoneCallURL)) {
            let alertController = UIAlertController(title: "MyApp", message: "Are you sure you want to call \n\(self.strPhoneNumber)?", preferredStyle: .alert)
            let yesPressed = UIAlertAction(title: "Yes", style: .default, handler: { (action) in
                application.openURL(phoneCallURL)
            })
            let noPressed = UIAlertAction(title: "No", style: .default, handler: { (action) in

            })
            alertController.addAction(yesPressed)
            alertController.addAction(noPressed)
            present(alertController, animated: true, completion: nil)
        }
    }

Answer 6

错误地我的回答是错误的,请查看这个:你可以使用这个:

guard let url = URL(string: "tel://\(yourNumber)") else {
return //be safe
}

if #available(iOS 10.0, *) {
UIApplication.shared.open(url)
} else {
UIApplication.shared.openURL(url)
}

我们需要检查我们是否在iOS 10或更高版本上因为'openURL'在iOS 10.0中已被弃用