How to use openURL for making a phone call in Swift?

问题

I have converted the code for making a phone call from Objective-C to Swift, but in Objective-C, we can set the type of the URL that we like to open (e.g. telephone, SMS, web) like this:

@"tel:xx"
@"mailto:info@example.es"
@"http://stackoverflow.com"
@"sms:768number"

The code in Swift is:

UIApplication.sharedApplication().openURL(NSURL(string : "9809088798")

I read that have not released any scheme parameter for tel:, but I don't know if Swift can detect if the string is for making a phone call, sending email, or opening a website. Or may I write:

(string : "tel//:9809088798")

?

回答1:

I am pretty sure you want:

UIApplication.sharedApplication().openURL(NSURL(string: "tel://9809088798")!)

(note that in your question text, you put tel//:, not tel://).

NSURL's string init expects a well-formed URL. It will not turn a bunch of numbers into a telephone number. You sometimes see phone-number detection in UIWebView, but that's being done at a higher level than NSURL.

EDIT: Added ! per comment below

回答2:

A self-contained solution in Swift:

private func callNumber(phoneNumber:String) {
  if let phoneCallURL:NSURL = NSURL(string:"tel://\(phoneNumber)") {
    let application:UIApplication = UIApplication.sharedApplication()
    if (application.canOpenURL(phoneCallURL)) {
      application.openURL(phoneCallURL);
    }
  }
}

Now, you should be able to use callNumber("7178881234") to make a call.

回答3:

For Swift in iOS:

var url:NSURL? = NSURL(string: "tel://9809088798")
UIApplication.sharedApplication().openURL(url!)

回答4:

You need to remember to remove the whitespaces or it won't work:

if let telephoneURL = NSURL(string: "telprompt://\(phoneNumber.stringByReplacingOccurrencesOfString(" ", withString: ""))") {
        UIApplication.sharedApplication().openURL(telelphoneURL)
    }

"telprompt://" will prompt the user to call or cancel while "tel://" will call directly.

回答5:

@ confile:

The problem is that your solution does not return to the app after the phone call has been finished on iOS7. – Jun 19 at 13:50

&@ Zorayr

Hm, curious if there is a solution that does do that.. might be a restriction on iOS.

use

UIApplication.sharedApplication().openURL(NSURL(string: "telprompt://9809088798")!)

You will get a prompt to Call/Cancel but it returns to your application. AFAIK there is no way to return (without prompting)

回答6:

You must insert "+"\ is another way

private func callNumber(phoneNumber:String) {
  if let phoneCallURL:NSURL = NSURL(string:"tel://"+"\(phoneNumber)") {
    let application:UIApplication = UIApplication.sharedApplication()
    if (application.canOpenURL(phoneCallURL)) {
      application.openURL(phoneCallURL);
    }
  }
}

回答7:

Small update to Swift 3

UIApplication.shared.openURL(NSURL(string: "telprompt://9809088798")! as URL)

回答8:

The following code snippet can tell if the SIM is there or not and if the device is capable of making the call and if ok then it'll make the call

  var info = CTTelephonyNetworkInfo()
    var carrier = info.subscriberCellularProvider
    if carrier != nil && carrier.mobileNetworkCode == nil || carrier.mobileNetworkCode.isEqual("") {
       //SIM is not there in the phone
    }
    else if UIApplication.sharedApplication().canopenURL(NSURL(string: "tel://9809088798")!)
    {
    UIApplication.sharedApplication().openURL(NSURL(string: "tel://9809088798")!)
    }
    else    
    {
      //Device does not have call making capability  
    }

回答9:

For swift 3

if let phoneCallURL:URL = URL(string:"tel://\(phoneNumber ?? "")") {
            let application:UIApplication = UIApplication.shared
            if (application.canOpenURL(phoneCallURL)) {
                application.open(phoneCallURL, options: [:], completionHandler: nil);
            }
        }

回答10:

For swift 4:

func call(phoneNumber: String) {
    if let url = URL(string: phoneNumber) {
        if #available(iOS 10, *) {
            UIApplication.shared.open(url, options: [:],
                                      completionHandler: {
                                        (success) in
                                        print("Open \(phoneNumber): \(success)")
            })
        } else {
            let success = UIApplication.shared.openURL(url)
            print("Open \(phoneNumber): \(success)")
        }
    }
} 

Then, use the function:

let phoneNumber = "tel://+132342424"
call(phoneNumber: phoneNumber)

回答11:

Same Answer in Swift 4.2

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 
   }
}

Call this like below

dialNumber(number: "+921111111222")

来源：https://stackoverflow.com/questions/24251259/how-to-use-openurl-for-making-a-phone-call-in-swift