快速拨打电话号码-Java 学习之路

我正在尝试拨打一个不使用特定号码的号码，而是拨打变量中的号码，或者至少告诉它拨打手机中的号码 . 在变量中调用的这个数字是我使用解析器或从网站sql中获取的数字 . 我做了一个按钮试图用功能调用存储在变量中的电话号码，但无济于事 . 什么都有帮助谢谢！

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)!)

        }

18 回答

试一试：

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

假设电话号码在 busPhone .

NSURL 的 init(string:) 返回一个Optional，所以通过使用 if let ，我们确保 url 是 NSURL （而不是 init 返回的 NSURL? ） .

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)
    }
}

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中已弃用

回复于 2024-05-03T18:16:19+08:00

iOS 10中的自包含解决方案， 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)
    }
  }
}

您应该可以使用 callNumber("7178881234") 拨打电话 .

回复于 2024-05-03T18:16:19+08:00

Swift 3.0和ios 10或更早版本

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)
        }
    }
}

回复于 2024-05-03T18:16:19+08:00

好的，我得到了帮助并想出来了 . 此外，我还放了一个漂亮的小警报系统，万一电话号码无效 . 我的问题是我说它是正确的，但数字有空格和不需要的字符，如（“123 456-7890”） . 如果您的号码是（“1234567890”），则UIApplication仅适用或接受 . 因此，您基本上通过创建一个新变量来仅删除数字来删除空格和无效字符 . 然后使用UIApplication调用这些数字 .

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()
        }
        }

回复于 2024-05-03T18:16:19+08:00

3
以上答案部分正确，但“tel：//”只有一个问题 . 通话结束后，它将返回主屏幕，而不是我们的应用程序 . 所以最好使用“telprompt：//”，它将返回到应用程序 .
```
var url:NSURL = NSURL(string: "telprompt://1234567891")!
UIApplication.sharedApplication().openURL(url)
```
回复于 2024-05-03T18:16:19+08:00

我在我的应用程序中使用此方法，它工作正常 . 我希望这对你也有帮助 .

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)
}

回复于 2024-05-03T18:16:19+08:00

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)
            }
        }
  }

回复于 2024-05-03T18:16:19+08:00

在Swift 3中，

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

回复于 2024-05-03T18:16:19+08:00

这是使用Swift 2.0更新@ Tom的答案注 - 这是我正在使用的整个CallComposer类 .

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
 }
}

回复于 2024-05-03T18:16:19+08:00

我正在使用swift 3解决方案进行数字验证

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)")!)
    }

回复于 2024-05-03T18:16:19+08:00

145

openURL（）已在iOS 10中弃用 . 以下是新语法：

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

回复于 2024-05-03T18:16:19+08:00

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.")
}

回复于 2024-05-03T18:16:19+08:00

斯威夫特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)

            }
        }
    }
}

回复于 2024-05-03T18:16:19+08:00

Swift 3.0解决方案：

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)

回复于 2024-05-03T18:16:19+08:00

这是使用 Scanner 将电话号码减少为有效组件的另一种方法...

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

回复于 2024-05-03T18:16:19+08:00

1

Swift 3.0 & iOS 10+

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

选项和完成处理程序是可选的，呈现：

UIApplication.shared.open(url)

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

回复于 2024-05-03T18:16:19+08:00

对于Swift 3.1和向后兼容的方法，请执行以下操作：

@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)
   }
}

回复于 2024-05-03T18:16:19+08:00

如果您的电话号码包含空格，请先将其删除！然后你可以使用accepted answer's解决方案 .

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)
    }
}

回复于 2024-05-03T18:16:19+08:00

快速拨打电话号码

18 回答

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