Question

我有一个保存电话号码的数据库（固定电话或手机或两者）。根据记录提取，如果用户想要拨打号码并且记录同时包含号码，移动电话和固定电话，如何给用户选择号码。

 @IBAction func makeCall(_ sender: Any) {
    print ("------Phone Number-----")
    print(landline)
    print(phoneNumber)

    let phone = "tel://";
    let lline = landline
    let url:NSURL = NSURL(string:phone+phoneNumber)!
    let url2:NSURL = NSURL(string: phone+landline)!
    UIApplication.shared.openURL(url as URL)
    UIApplication.shared.openURL(url2 as URL)
}

使用上面的代码，它会拨打这两个号码，但是想让用户选择其中一个号码。

Answer 1

您可以使用UIAlertController为用户提供选择：

@IBAction func makeCall(_ sender: Any) {
  print ("------Phone Number-----")
  print(landline)
  print(phoneNumber)

  let phone = "tel://";
  let lline = landline
  let url:NSURL = NSURL(string:phone+phoneNumber)!
  let url2:NSURL = NSURL(string: phone+landline)!

  let alert = UIAlertController(title: "Choose a number to call", message: "Please choose which number you want to call", preferredStyle: .alert)

  let firstNumberAction = UIAlertAction(title: "Number 1", style: .default) { (action) in
     UIApplication.shared.openURL(url as URL)
  }

  let secondNumberAction = UIAlertAction(title: "Number 2", style: .default) { (action) in
     UIApplication.shared.openURL(url2 as URL)
  }    

  alert.addAction(firstNumberAction)
  alert.addAction(secondNumberAction)
  self.present(alert, animated: true)
}

如果您愿意，您当然也可以添加取消操作。

Answer 2

作为weissja19的答案，你应该在调用openurl调用之前添加一个canopenurl检查。如果用户没有看到他无法执行的操作，也更整洁。我建议你使用这样的代码。

@IBAction func makeCall(_ sender: Any) {
    print ("------Phone Number-----")
    print(landline)
    print(phoneNumber)

    let phone = "tel://";
    let lline = landline
    let url:NSURL = NSURL(string:phone+phoneNumber)!
    let url2:NSURL = NSURL(string: phone+landline)!

    let alert = UIAlertController(title: 'Choose a number to call', message: 'Please choose which number you want to call', preferredStyle: .alert)
    if UIApplication.shared.canOpenUrl(url as URL) {
        let firstNumberAction = UIAlertAction(title: "Number 1", style: .default, handler: { _in
             UIApplication.shared.openURL(url as URL)
        })
        alert.addAction(firstNumberAction)
    }

    if UIApplication.shared.canOpenUrl(url2 as URL) {
        let secondNumberAction = UIAlertAction(title: "Number 2", style: .default, handler: { _in
             UIApplication.shared.openURL(url2 as URL)
        })
        alert.addAction(secondNumberAction)
    }    
    if alert.actions.count == 0 {
        alert.title = "No numbers to call"
        alert.message = ""
        alert.addAction(UIAlertAction(title: "OK", style: .default, handler: nil))
    } else {
        alert.addAction(UIAlertAction(title: "Cancel", style: .destructive, handler: nil))
    }
    self.presentViewController(alert, animated: true, completion: nil)
}

Answer 3

这两个代码都有效。以下是工作代码

set sftp:connect-program "ssh -ax -i key-file"

非常感谢weissja19和Ben Ong

Swift 3：如何选择拨打电话号码

3 个答案: