프로그래밍 방식으로 전화 걸기

프로그래밍 방식으로 전화 걸기

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iPhone에서 프로그래밍 방식으로 전화를 걸려면 어떻게해야합니까? 다음 코드를 시도했지만 아무 일도 일어나지 않았습니다.

NSString *phoneNumber = mymobileNO.titleLabel.text;
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];

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아마도 mymobileNO.titleLabel.text 값에 tel : // 체계가 포함되어 있지 않습니다 .

코드는 다음과 같아야합니다.

NSString *phoneNumber = [@"tel://" stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];

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원래 앱으로 돌아가려면 tel : // 대신 telprompt : //를 사용할 수 있습니다. tell prompt는 먼저 사용자에게 프롬프트하지만 전화가 끝나면 앱으로 돌아갑니다.

NSString *phoneNumber = [@"telprompt://" stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];

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@Cristian Radu 및 @Craig Mellon의 답변과 @ joel.d의 주석을 병합하면 다음을 수행해야합니다.

NSURL *urlOption1 = [NSURL URLWithString:[@"telprompt://" stringByAppendingString:phone]];
NSURL *urlOption2 = [NSURL URLWithString:[@"tel://" stringByAppendingString:phone]];
NSURL *targetURL = nil;

if ([UIApplication.sharedApplication canOpenURL:urlOption1]) {
    targetURL = urlOption1;
} else if ([UIApplication.sharedApplication canOpenURL:urlOption2]) {
    targetURL = urlOption2;
}

if (targetURL) {
    if (@available(iOS 10.0, *)) {
        [UIApplication.sharedApplication openURL:targetURL options:@{} completionHandler:nil];
    } else {
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wdeprecated-declarations"
        [UIApplication.sharedApplication openURL:targetURL];
#pragma clang diagnostic pop
    }
} 

먼저 "telprompt : //"URL을 사용하려고 시도하고 실패하면 "tel : //"URL을 사용합니다. 두 가지 모두 실패하면 iPad 또는 iPod Touch에서 전화를 겁니다.

스위프트 버전 :

let phone = mymobileNO.titleLabel.text
let phoneUrl = URL(string: "telprompt://\(phone)"
let phoneFallbackUrl = URL(string: "tel://\(phone)"
if(phoneUrl != nil && UIApplication.shared.canOpenUrl(phoneUrl!)) {
  UIApplication.shared.open(phoneUrl!, options:[String:Any]()) { (success) in
    if(!success) {
      // Show an error message: Failed opening the url
    }
  }
} else if(phoneFallbackUrl != nil && UIApplication.shared.canOpenUrl(phoneFallbackUrl!)) {
  UIApplication.shared.open(phoneFallbackUrl!, options:[String:Any]()) { (success) in
    if(!success) {
      // Show an error message: Failed opening the url
    }
  }
} else {
    // Show an error message: Your device can not do phone calls.
}

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The answers here are perfectly working. I am just converting Craig Mellon answer to Swift. If someone comes looking for swift answer, this will help them.

 var phoneNumber: String = "telprompt://".stringByAppendingString(titleLabel.text!) // titleLabel.text has the phone number.
        UIApplication.sharedApplication().openURL(NSURL(string:phoneNumber)!)

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If you are using Xamarin to develop an iOS application, here is the C# equivalent to make a phone call within your application:

string phoneNumber = "1231231234";
NSUrl url = new NSUrl(string.Format(@"telprompt://{0}", phoneNumber));
UIApplication.SharedApplication.OpenUrl(url);

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Swift 3

let phoneNumber: String = "tel://3124235234"
UIApplication.shared.openURL(URL(string: phoneNumber)!)

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In Swift 3.0,

static func callToNumber(number:String) {

        let phoneFallback = "telprompt://\(number)"
        let fallbackURl = URL(string:phoneFallback)!

        let phone = "tel://\(number)"
        let url = URL(string:phone)!

        let shared = UIApplication.shared

        if(shared.canOpenURL(fallbackURl)){
            shared.openURL(fallbackURl)
        }else if (shared.canOpenURL(url)){
            shared.openURL(url)
        }else{
            print("unable to open url for call")
        }

    }

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The Java RoboVM equivalent:

public void dial(String number)
{
  NSURL url = new NSURL("tel://" + number);
  UIApplication.getSharedApplication().openURL(url);
}

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Tried the Swift 3 option above, but it didnt work. I think you need the following if you are to run against iOS 10+ on Swift 3:

Swift 3 (iOS 10+):

let phoneNumber = mymobileNO.titleLabel.text       
UIApplication.shared.open(URL(string: phoneNumber)!, options: [:], completionHandler: nil)

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let phone = "tel://\("1234567890")";
let url:NSURL = NSURL(string:phone)!;
UIApplication.sharedApplication().openURL(url);

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'openURL:' is deprecated: first deprecated in iOS 10.0 - Please use openURL:options:completionHandler: instead

in Objective-c iOS 10+ use :

NSString *phoneNumber = [@"tel://" stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber] options:@{} completionHandler:nil];

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Swift

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

참고URL : https://stackoverflow.com/questions/4929717/make-a-phone-call-programmatically