如何在iPhone上以编程方式拨打电话？我尝试了以下代码，但没有任何反应：

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

mymobileNO.titleLabel.text值可能不包含方案tel：//

您的代码应如下所示：

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

要返回原始应用，您可以使用telprompt：//而不是tel：//-提示提示符将首先提示用户，但是当通话结束时，它将返回您的应用：

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

合并@Cristian Radu和@Craig Mellon的答案以及@ joel.d的评论，您应该执行以下操作：

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
    }
} 

这将首先尝试使用“ telprompt：//” URL，如果失败，它将使用“ tel：//” URL。如果两者均失败，则您尝试在iPad或iPod Touch上拨打电话。

迅捷版本：

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.
}

这里的答案是完美的工作。我只是将Craig Mellon的答案转换为Swift。如果有人来寻找快速答案，这将对他们有所帮助。

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

如果您使用Xamarin开发iOS应用程序，则在您的应用程序中拨打电话等效于C＃：

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

迅捷3

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

在Swift 3.0中，

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")
        }

    }

Java RoboVM等效项：

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

尝试了上面的Swift 3选项，但是没有用。如果要在Swift 3上针对iOS 10+运行，我认为您需要以下条件：

Swift 3（iOS 10以上）：

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

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

迅速

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

不建议使用“ openURL：”：在iOS 10.0中首先不建议使用-请改用openURL：options：completionHandler：

在Objective-c iOS 10+中使用：

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