ios 以编程方式拨打电话

Make a phone call programmatically

提问by user564963

How can I make a phone call programmatically on iPhone? I tried the following code but nothing happened:

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]];

回答by Craig Mellon

To go back to original app you can use telprompt:// instead of tel:// - The tell prompt will prompt the user first, but when the call is finished it will go back to your app:

要返回原始应用程序，您可以使用 telprompt:// 而不是 tel:// - tell 提示将首先提示用户，但当呼叫完成后，它将返回到您的应用程序：

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

回答by Cristian Radu

Probably the mymobileNO.titleLabel.textvalue doesn't include the scheme tel://

可能mymobileNO.titleLabel.text值不包括方案tel://

Your code should look like this:

您的代码应如下所示：

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

回答by Guillaume Boudreau

Merging the answers of @Cristian Radu and @Craig Mellon, and the comment from @joel.d, you should do:

合并@Cristian Radu 和@Craig Mellon 的答案，以及@joel.d 的评论，你应该这样做：

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
    }
} 

This will first try to use the "telprompt://" URL, and if that fails, it will use the "tel://" URL. If both fails, you're trying to place a phone call on an iPad or iPod Touch.

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

Swift Version :

快速版本：

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.
}

回答by Shivaay

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.

这里的答案非常有效。我只是将 Craig Mellon 的答案转换为 Swift。如果有人来寻求快速答复，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)!)

回答by Michael Kniskern

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

如果您使用 Xamarin 开发 iOS 应用程序，这里是在您的应用程序中拨打电话的等效 C#：

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

回答by Jorge Costa

Swift 3

斯威夫特 3

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

回答by Rohit Pathak

In Swift 3.0,

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")
        }

    }

回答by Charlie Seligman

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:

尝试了上面的 Swift 3 选项，但没有用。如果您要在 Swift 3 上针对 iOS 10+ 运行，我认为您需要以下内容：

Swift 3 (iOS 10+):

斯威夫特 3（iOS 10+）：

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

回答by EntangledLoops

The Java RoboVM equivalent:

Java RoboVM 等效项：

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

回答by Himali Shah

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